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∫u dv = uv −
∫v du
± diff. int. (±∫
)
+ u dv (∫u dv)
↘− du v (−
∫v du)
A Lecture on Row Integration by Parts (RIP)
John A. Rock
Loyola Marymountπ.2017
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 1 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Integration by Parts
We can iterate “ultra-violet voodoo”:∫u dv = uv −
∫v du
= u1v1−∫v1 du1
= u1v1 − u2v2 +∫v2 du2
= u1v1 − u2v2 + u3v3−∫v3 du3
...
. . . but when do we stop?
It completely depends on the last integral (−1)n∫vn dun.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 2 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2
(+∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2?
Not yet. . .)↘
− u4 v3 (−∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Tabular Row Integration by Parts (RIP)
Given∫u dv = uv −
∫v du, define
{u1 = u, v1 =
∫v0 =
∫dv ,
un+1 = u′n, vn+1 =∫vn.
± diff. int. (±∫
)
+ u1 v0 (The top row is the original∫u dv .)
↘− u2 v1 (−
∫u2v1 = −
∫v du = −
∫v1 du1.)
↘+ u3 v2 (+
∫v2 du2? Not yet. . .)
↘− u4 v3 (−
∫v3 du3? Stop!)
∫u dv = u1v1 − u2v2 + u3v3−
∫v3 du3.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 3 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x
↘− 2x − cos x (
∫2x cos x? Keep going. . .)
↘+ 2 − sin x (−
∫2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x
(∫
2x cos x? Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x?
Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)
↘+ 2 − sin x (−
∫2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)↘
+ 2 − sin x
(−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x
= 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1.∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x (∫
2x cos x? Keep going. . .)↘
+ 2 − sin x (−∫
2 sin x = 2 cos x + C . Stop!)
∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 4 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1 (as in the film).∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x↘
+ 2 − sin x↘
− 0 cos x (−∫
0 = C .)
∫x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 5 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1 (as in the film).∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x↘
+ 2 − sin x↘
− 0 cos x
(−∫
0 = C .)
∫x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 5 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1 (as in the film).∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x↘
+ 2 − sin x↘
− 0 cos x (−∫
0 = C .)
∫x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 5 / 21
Stand and Deliver “Tic-Tac-Toe”
Example 1 (as in the film).∫x2 sin x dx
± diff. int. (±∫
)
+ x2 sin x↘
− 2x − cos x↘
+ 2 − sin x↘
− 0 cos x (−∫
0 = C .)
∫x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 5 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x↘
− 1/x x2/2 (−(1/2)∫x = −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x
↘− 1/x x2/2 (−(1/2)
∫x = −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x↘
− 1/x x2/2
(−(1/2)∫x = −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x↘
− 1/x x2/2 (−(1/2)∫x
= −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x↘
− 1/x x2/2 (−(1/2)∫x = −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
LIPET, LIATE, etc.
Example 2.∫x ln x dx
± diff. int. (±∫
)
+ ln x x↘
− 1/x x2/2 (−(1/2)∫x = −x2/4 + C , stop.)
∫x ln x dx = (1/2)x2 ln x −x2/4 + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 6 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x
↘− 2e2x (sin 3x)/3 (−(2/3)
∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3
(−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x?
Try again.)↘
+ 4e2x (− cos 3x)/9 (−(4/9)∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9
(−(4/9)∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!
∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .
∫e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
An example without polynomials
Example 3.∫e2x cos 3x dx
± diff. int. (±∫
)
+ e2x cos 3x↘
− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)
↘+ 4e2x (− cos 3x)/9 (−(4/9)
∫e2x cos 3x , stop.)
The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)
∫e2x cos 3x+C .∫
e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 7 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3
(−∫
2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)
↘+ 2/x x3/3 (+
∫2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3
(+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3
= 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
One more example
Example 4.∫
3x2 ln2 x dx
± diff. int. (±∫
)
+ ln2 x 3x2
↘− (2 ln x)/x x3 (−
∫2x2 ln x dx)
− 2 ln x x2 ← (−∫
2x2 ln x dx , same as above.)↘
+ 2/x x3/3 (+∫
2x2/3 = 2x3/9 + C )
∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 8 / 21
Your turn!
Exercises. Find the following antiderivatives.
1∫
(x3 − 2x + 7)ex dx
2∫
sin 2t sin 5t dt
3∫
cos x ln (sin x) dx
4∫
ln (x2 + 1) dx
5∫
(arcsin x)2 dx
(For fun, let’s see what WolframAlpha or Symbolab gives us.)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 9 / 21
Your turn!
Exercises. Find the following antiderivatives.
1∫
(x3 − 2x + 7)ex dx
2∫
sin 2t sin 5t dt
3∫
cos x ln (sin x) dx
4∫
ln (x2 + 1) dx
5∫
(arcsin x)2 dx
(For fun, let’s see what WolframAlpha or Symbolab gives us.)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 9 / 21
Solutions
Exercise 1.∫
(x3 − 2x + 7)ex dx
± diff. int. (±∫
)
+ x3 − 2x + 7 ex
↘− 3x2 − 2 ex
↘+ 6x ex
↘− 6 ex
↘+ 0 ex (
∫0 = C )
∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 10 / 21
Solutions
Exercise 1.∫
(x3 − 2x + 7)ex dx
± diff. int. (±∫
)
+ x3 − 2x + 7 ex
↘− 3x2 − 2 ex
↘+ 6x ex
↘− 6 ex
↘+ 0 ex (
∫0 = C )
∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 10 / 21
Solutions
Exercise 1.∫
(x3 − 2x + 7)ex dx
± diff. int. (±∫
)
+ x3 − 2x + 7 ex
↘− 3x2 − 2 ex
↘+ 6x ex
↘− 6 ex
↘+ 0 ex
(∫
0 = C )
∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 10 / 21
Solutions
Exercise 1.∫
(x3 − 2x + 7)ex dx
± diff. int. (±∫
)
+ x3 − 2x + 7 ex
↘− 3x2 − 2 ex
↘+ 6x ex
↘− 6 ex
↘+ 0 ex (
∫0 = C )
∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 10 / 21
Solutions
Exercise 1.∫
(x3 − 2x + 7)ex dx
± diff. int. (±∫
)
+ x3 − 2x + 7 ex
↘− 3x2 − 2 ex
↘+ 6x ex
↘− 6 ex
↘+ 0 ex (
∫0 = C )
∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 10 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t
↘− 2 cos 2t (− cos 5t)/5
↘+ −4 sin 2t (− sin 5t)/25 (+(4/25)
∫sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5
↘+ −4 sin 2t (− sin 5t)/25 (+(4/25)
∫sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25
(+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!
∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 2.∫
sin 2t sin 5t dt
± diff. int. (±∫
)
+ sin 2t sin 5t↘
− 2 cos 2t (− cos 5t)/5↘
+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫
sin 2t sin 5t dt)
The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t
+(4/25)∫
sin 2t sin 5t dt + C
∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 11 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x↘
− cos x
sin xsin x (−
∫cos x = − sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x
↘− cos x
sin xsin x (−
∫cos x = − sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x↘
− cos x
sin xsin x
(−∫
cos x = − sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x↘
− cos x
sin xsin x (−
∫cos x =
− sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x↘
− cos x
sin xsin x (−
∫cos x = − sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 3.∫
cos x ln (sin x) dx
± diff. int. (±∫
)
+ ln (sin x) cos x↘
− cos x
sin xsin x (−
∫cos x = − sin x + C )
∫cos x ln sin x dx = sin x ln (sin x)− sin x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 12 / 21
Solutions
Exercise 4.∫
ln (x2 + 1) dx
± diff. int. (±∫
)
+ ln (x2 + 1) 1↘
− 2x/(x2 + 1) x (−∫
2x2/(x2 + 1))
∫ln (x2 + 1) dx = x ln (x2 + 1)−
∫2x2/(x2 + 1)
= x ln (x2 + 1)−∫
2(x2 + 1− 1)/(x2 + 1)
= x ln (x2 + 1)−∫
2 +∫
2/(x2 + 1)
= x ln (x2 + 1)−2x + 2 arctan x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 13 / 21
Solutions
Exercise 4.∫
ln (x2 + 1) dx
± diff. int. (±∫
)
+ ln (x2 + 1) 1
↘− 2x/(x2 + 1) x (−
∫2x2/(x2 + 1))
∫ln (x2 + 1) dx = x ln (x2 + 1)−
∫2x2/(x2 + 1)
= x ln (x2 + 1)−∫
2(x2 + 1− 1)/(x2 + 1)
= x ln (x2 + 1)−∫
2 +∫
2/(x2 + 1)
= x ln (x2 + 1)−2x + 2 arctan x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 13 / 21
Solutions
Exercise 4.∫
ln (x2 + 1) dx
± diff. int. (±∫
)
+ ln (x2 + 1) 1↘
− 2x/(x2 + 1) x
(−∫
2x2/(x2 + 1))
∫ln (x2 + 1) dx = x ln (x2 + 1)−
∫2x2/(x2 + 1)
= x ln (x2 + 1)−∫
2(x2 + 1− 1)/(x2 + 1)
= x ln (x2 + 1)−∫
2 +∫
2/(x2 + 1)
= x ln (x2 + 1)−2x + 2 arctan x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 13 / 21
Solutions
Exercise 4.∫
ln (x2 + 1) dx
± diff. int. (±∫
)
+ ln (x2 + 1) 1↘
− 2x/(x2 + 1) x (−∫
2x2/(x2 + 1))
∫ln (x2 + 1) dx = x ln (x2 + 1)−
∫2x2/(x2 + 1)
= x ln (x2 + 1)−∫
2(x2 + 1− 1)/(x2 + 1)
= x ln (x2 + 1)−∫
2 +∫
2/(x2 + 1)
= x ln (x2 + 1)−2x + 2 arctan x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 13 / 21
Solutions
Exercise 4.∫
ln (x2 + 1) dx
± diff. int. (±∫
)
+ ln (x2 + 1) 1↘
− 2x/(x2 + 1) x (−∫
2x2/(x2 + 1))
∫ln (x2 + 1) dx = x ln (x2 + 1)−
∫2x2/(x2 + 1)
= x ln (x2 + 1)−∫
2(x2 + 1− 1)/(x2 + 1)
= x ln (x2 + 1)−∫
2 +∫
2/(x2 + 1)
= x ln (x2 + 1)−2x + 2 arctan x + C .
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 13 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1
↘− 2 arcsin x√
1−x2 x (−∫
2x arcsin x√1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x
(−∫
2x arcsin x√1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2
(−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2
= −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Solutions
Exercise 5.∫
(arcsin x)2 dx
± diff. int. (±∫
)
+ (arcsin x)2 1↘
− 2 arcsin x√1−x2 x (−
∫2x arcsin x√
1−x2 )
− arcsin x 2x√1−x2 ← (−
∫2x arcsin x√
1−x2 )
↘+ 1√
1−x2 −2√
1− x2 (−∫
2 = −2x + C )
∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)
√1− x2−2x + C
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 14 / 21
Taylor’s Formula with Integral Remainder
Theorem 1 (Horowitz). Under suitable conditions,
f (b) = f (a) +f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
To derive this formula using RIP, there are two tricks:
1. Start with the Fundamental Theorem of Calculus:
f (b)− f (a) =∫ ba f ′(t) dt =
∫ ba (−1)(−f ′(t)) dt.
2. Use (b − t) as the antiderivative of (−1) (treating b as a constant).
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 15 / 21
Taylor’s Formula with Integral Remainder
Theorem 1 (Horowitz). Under suitable conditions,
f (b) = f (a) +f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
To derive this formula using RIP, there are two tricks:
1. Start with the Fundamental Theorem of Calculus:
f (b)− f (a) =∫ ba f ′(t) dt =
∫ ba (−1)(−f ′(t)) dt.
2. Use (b − t) as the antiderivative of (−1) (treating b as a constant).
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 15 / 21
Taylor’s Formula with Integral Remainder
Theorem 1 (Horowitz). Under suitable conditions,
f (b) = f (a) +f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
To derive this formula using RIP, there are two tricks:
1. Start with the Fundamental Theorem of Calculus:
f (b)− f (a) =∫ ba f ′(t) dt =
∫ ba (−1)(−f ′(t)) dt.
2. Use (b − t) as the antiderivative of (−1) (treating b as a constant).
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 15 / 21
Taylor’s Formula with Integral Remainder
Theorem 1 (Horowitz). Under suitable conditions,
f (b) = f (a) +f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
To derive this formula using RIP, there are two tricks:
1. Start with the Fundamental Theorem of Calculus:
f (b)− f (a) =∫ ba f ′(t) dt =
∫ ba (−1)(−f ′(t)) dt.
2. Use (b − t) as the antiderivative of (−1) (treating b as a constant).
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 15 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1↘
− −f (2)(t) (b − t)↘
+ −f (3)(t) −(b − t)2
2!...
......
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1
↘− −f (2)(t) (b − t)
↘
+ −f (3)(t) −(b − t)2
2!...
......
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1↘
− −f (2)(t) (b − t)
↘
+ −f (3)(t) −(b − t)2
2!...
......
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1↘
− −f (2)(t) (b − t)↘
+ −f (3)(t) −(b − t)2
2!
......
...
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1↘
− −f (2)(t) (b − t)↘
+ −f (3)(t) −(b − t)2
2!...
......
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!
↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
± diff. int. (±∫
)
+ −f (1)(t) −1↘
− −f (2)(t) (b − t)↘
+ −f (3)(t) −(b − t)2
2!...
......
(−1)n−1 −f (n)(t) (−1)n(b − t)n−1
(n − 1)!↘
(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n
n!
(+∫ f (n+1)(t)
n! (b − t)n dt)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 16 / 21
Taylor’s Formula with Integral Remainder
Therefore,
f (b)−f (a) =
∫ b
a−f (1)(t)(−1) dt
=
[−f (1)(t)(b − t)− f (2)(t)
2!(b − t)2 − · · · − f (n)(t)
n!(b − t)n
]ba
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt
=f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 17 / 21
Taylor’s Formula with Integral Remainder
Therefore,
f (b)−f (a) =
∫ b
a−f (1)(t)(−1) dt
=
[−f (1)(t)(b − t)− f (2)(t)
2!(b − t)2 − · · · − f (n)(t)
n!(b − t)n
]ba
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt
=f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 17 / 21
Taylor’s Formula with Integral Remainder
Therefore,
f (b)−f (a) =
∫ b
a−f (1)(t)(−1) dt
=
[−f (1)(t)(b − t)− f (2)(t)
2!(b − t)2 − · · · − f (n)(t)
n!(b − t)n
]ba
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt
=f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 17 / 21
Taylor’s Formula with Integral Remainder
Therefore,
f (b)−f (a) =
∫ b
a−f (1)(t)(−1) dt
=
[−f (1)(t)(b − t)− f (2)(t)
2!(b − t)2 − · · · − f (n)(t)
n!(b − t)n
]ba
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt
=f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 17 / 21
Taylor’s Formula with Integral Remainder
Therefore,
f (b)−f (a) =
∫ b
a−f (1)(t)(−1) dt
=
[−f (1)(t)(b − t)− f (2)(t)
2!(b − t)2 − · · · − f (n)(t)
n!(b − t)n
]ba
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt
=f ′(a)
1!(b − a) +
f ′′(a)
2!(b − a)2 + · · ·+ f (n)(a)
n!(b − a)n
+
∫ b
a
f (n+1)(t)
n!(b − t)n dt.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 17 / 21
Laplace Transform of f (n)
Theorem 2 (Horowitz). Under suitable conditions,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0) + snL{f (t)}.
This formula follows immediately from RIP!
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 18 / 21
Laplace Transform of f (n)
Theorem 2 (Horowitz). Under suitable conditions,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0) + snL{f (t)}.
This formula follows immediately from RIP!
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 18 / 21
Laplace Transform of f (n)
± diff. int. (±∫
)
+ e−st f (n)(t)↘
− −se−st f (n−1)(t)↘
+ s2e−st f (n−2)(t)...
......
(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘
(−1)n (−1)nsne−st → f (t)(+sn
∫e−st f (t) dt
)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 19 / 21
Laplace Transform of f (n)
± diff. int. (±∫
)
+ e−st f (n)(t)
↘− −se−st f (n−1)(t)
↘+ s2e−st f (n−2)(t)...
......
(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘
(−1)n (−1)nsne−st → f (t)(+sn
∫e−st f (t) dt
)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 19 / 21
Laplace Transform of f (n)
± diff. int. (±∫
)
+ e−st f (n)(t)↘
− −se−st f (n−1)(t)↘
+ s2e−st f (n−2)(t)
......
...
(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘
(−1)n (−1)nsne−st → f (t)(+sn
∫e−st f (t) dt
)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 19 / 21
Laplace Transform of f (n)
± diff. int. (±∫
)
+ e−st f (n)(t)↘
− −se−st f (n−1)(t)↘
+ s2e−st f (n−2)(t)...
......
(−1)n−1 (−1)n−1sn−1e−st f (1)(t)
↘(−1)n (−1)nsne−st → f (t)
(+sn
∫e−st f (t) dt
)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 19 / 21
Laplace Transform of f (n)
± diff. int. (±∫
)
+ e−st f (n)(t)↘
− −se−st f (n−1)(t)↘
+ s2e−st f (n−2)(t)...
......
(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘
(−1)n (−1)nsne−st → f (t)(+sn
∫e−st f (t) dt
)
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 19 / 21
Laplace Transform of f (n)
Therefore,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)
]∞0
+ sn∫ ∞0
e−st f (t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)
+ snL{f (t)}.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 20 / 21
Laplace Transform of f (n)
Therefore,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)
]∞0
+ sn∫ ∞0
e−st f (t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)
+ snL{f (t)}.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 20 / 21
Laplace Transform of f (n)
Therefore,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)
]∞0
+ sn∫ ∞0
e−st f (t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)
+ snL{f (t)}.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 20 / 21
Laplace Transform of f (n)
Therefore,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)
]∞0
+ sn∫ ∞0
e−st f (t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)
+ snL{f (t)}.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 20 / 21
Laplace Transform of f (n)
Therefore,
L{f (n)(t)} =
∫ ∞0
e−st f (n)(t) dt
=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)
]∞0
+ sn∫ ∞0
e−st f (t) dt
=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)
+ snL{f (t)}.
John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 20 / 21