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MTH131 Review Integration Worksheet Part 1
!"#$
%− !'#$
(+ 𝑐, = 3 .
!𝑑𝑥 = 3 ∗ lnx + 𝐶 = −1 sin 𝑥 𝑑𝑥
= −[− cos 𝑥 ] + 𝑐 = cos(x) +C Use trig rule 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑟𝑢𝑙𝑒 𝑥F = !G#$
FH. 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑟𝑢𝑙𝑒 𝑏𝑟𝑖𝑛𝑔 contstants out
and 1𝑥 = 𝐼𝑛 𝑥
u = 2x, !$S#$
TS+ !U
$"#$
'"
+ 𝐶 u= 3x and du=3 dx .
(𝑑𝑢 = 𝑑𝑥 → 𝑒W .
( du
du = 2 dx 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑟𝑢𝑙𝑒 𝑥F = !G#$
FH. .
(𝑒W 𝑑𝑢 = .
(𝑒W + 𝐶
.X𝑑𝑢 = 𝑑𝑥 → cos 𝑢 .
X𝑑𝑢 = .
(𝑒(! + 𝐶
½ sin u+ C = ½ sin (2x) + c 𝑈𝑠𝑒 𝑒! = 𝑒! 𝑟𝑢𝑙𝑒 Use u substitution and trig identity
= 𝑡𝑎𝑛Z. 𝑥 2 .
.Z!'𝑑𝑥 = 2 ∗ 𝑠𝑖𝑛Z. 𝑥 + 𝐶 = 3! .[
(+ 𝐶
Trig identity Trig Identity May not need to know this #7 and # 8 use rule 𝑎! = \]
^[\+C
Part 2 – U substitution
u= 7x+2 u=2𝒙𝟐 u=𝒙𝟑 du= 7dx , 𝟏
𝟕𝒅𝒖 = 𝒅𝒙 du= 4x, 𝟏
𝟒𝒅𝒖 = 𝒙 ∗ 𝒅𝒙 du=3𝒙𝟐dx
𝒔𝒊𝒏(𝒖) 𝟏𝟕 𝒅𝒖 = 𝟏
𝟕𝒔𝒊𝒏(𝒖)𝒅𝒖 = 𝟕 𝒙𝒆𝟐𝒙𝟐dx = 𝒄𝒐𝒔 𝒖 𝒅𝒖
𝟏𝟕−𝒄𝒐𝒔𝒖 + 𝒄 =𝟕 𝒆𝒖 𝟏
𝟒𝒅𝒖 = 𝟕
𝟒𝒆𝒖𝒅𝒖 =[sinu] + C = sin(𝒙𝟑) + 𝑪
= −𝟏𝟕𝒄𝒐𝒔 𝟕𝒙 + 𝟐 + 𝑪 = 𝟕
𝟒𝒆𝟐𝒙𝟐 + 𝑪 Use trig rule
Use substitution and trig identity use 𝑒! = 𝑒! 𝑟𝑢𝑙𝑒
u= ln x u= ln x u=7𝒙𝟐+1 du= 1/x dx du= 1/x dx du= 14x dx 𝟏
𝟏𝟒𝒅𝒖 = 𝒙 ∗ 𝒅𝒙
..HW'
𝑑𝑢 = tan 𝑢 Z. + 𝐶 u du = W'
X+ 𝐶 = 𝟑 𝒙( 𝒙𝟐+1)𝟖𝒅𝒙
tan lnx Z.+C = qrs'
X +C =3 𝑢t .
.%𝑑𝑢 = (
.%𝑢t𝑑𝑢
Trig identity, Rule 𝑥𝑟 =𝑥𝑟+1
𝑟+1 = (
.%∗ Wu
v+ 𝑐 → .
%X(𝒙𝟐+1)𝟗+C
may not need to know Great free resources to check your answers: http://www.wolframalpha.com/input/?i=integral+3x*%287x%5E2%2B1%29%5E8+dx For more problem sets to practise a good website is: https://www.math.ucdavis.edu/~kouba/ProblemsList.html Helpful Rules and Laws when doing Integrals: Laws
∫x dx = x2/2 + C
∫x2 dx = x3/3 + C ∫(1/x) dx = ln|x| + C ∫ex dx = ex + C ∫ax dx = ax/ln(a) + C ∫ln(x) dx = x ln(x) − x + C ∫cos(x) dx = sin(x) + C ∫sin(x) dx = -cos(x) + C ∫sec2(x) dx = tan(x) + C
Rules
∫cf(x) dx = c ∫f(x) dx ∫xn dx = xn+1/(n+1) + C ∫(f + g) dx= ∫f dx + ∫g dx ∫(f - g) dx= ∫f dx - ∫g dx Integration by Parts When two functions are multiplied together, but is also helpful in other ways. You will see plenty of examples soon, but first let us see the rule:
∫u* dv = u*v −∫v du u is the function u(x) v is the function v(x) Ex. ∫ex sin(x) dx Choose u and v: u = sin(x), du= cos(x) dv = ex , v = ex
∫ex sin(x) dx = sin(x) ex -∫cos(x) ex dx (color code it to the integration by parts above) We can use integration AGAIN by parts again for ∫cos(x) ex dx: u = cos(x), du= -sin(x), v = ex dv = ex
∫u* dv = u*v −∫v du (plug and chug) ∫cos(x) ex dx= cos(x) ex + ∫sin(x) ex dx Now put it together: ∫ex sin(x) dx = sin(x) ex - (cos(x) ex −∫−sin(x) ex dx) Simplify: ∫ex sin(x) dx = ex sin(x) - ex cos(x) −∫ ex sin(x)dx Now we have the same integral on both sides (except one is subtracted) so bring ∫ ex sin(x)dx to one side ... 2∫ex sin(x) dx = ex sin(x) − ex cos(x) Simplify: ∫ex sin(x) dx = ex (sin(x) - cos(x)) / 2 + C