Solutions to U-Substitution

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SOLUTIONS TO U-SUBSTITUTION

SOLUTION 8 : Integrate $\displaystyle{ \int { 3x+6 \over x^2+4x-3 } \,dx }$ . Let

u = x²+4x-3

so that

du = (2x+4) dx = 2 (x+2) dx ,

(1/2) du = (x+2) dx .

Substitute into the original problem, replacing all forms of x, getting

$\displaystyle{ \int { 3x+6 \over x^2+4x-3 } \,dx } = \displaystyle{ \int { 3(x+2) \over x^2+4x-3 } \, dx }$

$= 3 \displaystyle{ \int { 1 \over x^2+4x-3 } \,(x+2) dx }$

$= 3 \displaystyle{ \int { 1 \over u } \, (1/2) du }$

$= (3/2) \displaystyle{ \int { 1 \over u } \, du }$

$= (3/2) \displaystyle{ { \ln \vert u\vert } + C }$

$= (3/2) \displaystyle{ { \ln \vert x^2+4x-3\vert } + C }$ .

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SOLUTION 9 : Integrate $\displaystyle{ \int { x \ 3^{ x^2+1} } \,dx }$ . Let

u = x²+1

so that

du = 2x dx ,

(1/2) du = x dx .

Substitute into the original problem, replacing all forms of x, getting

$\displaystyle{ \int { x \ 3^{ x^2+1} } \,dx } = \displaystyle{ \int { 3^{ x^2+1} } \, x dx }$

$= \displaystyle{ \int { 3^u } \,(1/2) du }$

$= (1/2) \displaystyle{ \int { 3^u } \, du }$

$= (1/2) \displaystyle{ { 3^u \over \ln 3 } + C }$

$= \displaystyle{ 1 \over 2 \ln 3 } \displaystyle{ { 3^u } + C }$

(Recall that $A \ln B = \ln B^A$ .)

$= \displaystyle{ 1 \over \ln 3^2 } \displaystyle{ { 3^{ x^2+1 } } + C }$

$= \displaystyle{ 1 \over \ln 9 } \displaystyle{ { 3^{ x^2+1 } } + C }$ .

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SOLUTION 10 : Integrate $\displaystyle{ \int { 3 \over x \ln x } \,dx }$ . Let

$u = \ln x$

so that

$du = \displaystyle{ 1 \over x } dx$ .

Substitute into the original problem, replacing all forms of x, getting

$\displaystyle{ \int { 3 \over x \ln x } \,dx } = 3 \displaystyle{ \int { 1 \over \ln x } \, { 1 \over x } dx }$

$= 3 \displaystyle{ \int { 1 \over u } \, du }$

$= 3 \displaystyle{ \ln \vert u \vert + C }$

$= 3 \displaystyle{ \ln \vert \ln x \vert + C }$ .

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SOLUTION 11 : Integrate $\displaystyle{ \int { \cos(5x) \over e^{ \sin(5x) } } \,dx }$ . Let

$u = \sin(5x)$

so that (Don't forget to use the chain rule.)

$du = 5 \cos(5x) dx$ ,

$(1/5) du = \cos(5x) dx$ .

Substitute into the original problem, replacing all forms of x, getting

$\displaystyle{ \int { \cos(5x) \over e^{ \sin(5x) } } \,dx } = \displaystyle{ \int { 1 \over e^{ \sin(5x) } } \, \cos(5x) dx }$

$= \displaystyle{ \int { 1 \over e^ u } \, (1/5) du }$

(Do not make the following VERY COMMON MISTAKE : $\displaystyle{ \int { 1 \over e^u } \, du } = \displaystyle{ \ln \vert e^u\vert } + C$ . Why is this INCORRECT ?)

$= (1/5) \displaystyle{ \int { e^{-u} } \, du }$ .

Now make another substitution. Let

w = -u

so that

dw = (-1) du ,

(-1) dw = du .

Substitute into the problem, replacing all forms of u, getting

$(1/5) \displaystyle{ \int { e^{-u} } \, du } = (1/5) \displaystyle{ \int { e^{w} } \, (-1) dw }$

$= (1/5)(-1) \displaystyle{ \int { e^{w} } \, dw }$

$= (-1/5) \displaystyle{ e^{w} + C }$

$= (-1/5) \displaystyle{ e^{-u} + C }$

$= (-1/5) \displaystyle{ e^{- \sin(5x) } + C }$ .

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SOLUTION 12 : Integrate $\displaystyle{ \int_{0}^{ \sqrt{\pi} } { x \sin(x^2) } \,dx }$ . Let

u = x²

so that

du = 2x dx ,

(1/2) du = x dx .

In addition, the range of x-values is

$x : 0 \longrightarrow \sqrt{ \pi }$ ,

so that the range of u-values is

$u : (0)^2 \longrightarrow \big( \sqrt{ \pi } \big)^2$ ,

$u : 0 \longrightarrow \pi$ .

Substitute into the original problem, replacing all forms of x and the x-values, getting

$\displaystyle{ \int_{0}^{ \sqrt{\pi} } { x \sin(x^2) } \,dx } = \displaystyle{ \int_{0}^{ \sqrt{\pi} } { \sin(x^2) } \, x dx }$

$= \displaystyle{ \int_{0}^{ \pi } { \sin(u) } \, (1/2) du }$

$= (1/2) \displaystyle{ \int_{0}^{ \pi } { \sin u } \, du }$

$= (1/2) \displaystyle{ \Big( -\cos u \big\vert_{0}^{\pi} \Big) }$

$= (-1/2) \displaystyle{ \Big( \cos u \big\vert_{0}^{\pi} \Big) }$

$= (-1/2)( \cos \pi - \cos 0 )$

= (-1/2)( (-1) - (1) )

= (-1/2)( -2)

= 1 .

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SOLUTION 13 : Integrate $\displaystyle{ \int { (x+3) (x-1)^5 } \,dx }$ . Let

u = x-1

so that

du = (1) dx = dx .

In addition, we can "back substitute" with

x = u+1 .

Substitute into the original problem, replacing all forms of x, getting

$\displaystyle{ \int { (x+3) (x-1)^5 } \,dx } = \displaystyle{ \int { ((u+1)+3) u^5 } \,du }$

$= \displaystyle{ \int { (u+4) u^5 } \,du }$

$= \displaystyle{ \int { (u^6 + 4 u^5) } \,du }$

$= \displaystyle{ { u^7 \over 7 } + 4 { u^6 \over 6 } + C }$

$= \displaystyle{ (1/7) u^7 + (2/3) u^6 + C }$

$= \displaystyle{ (1/7) (x-1)^7 + (2/3) (x-1)^6 + C }$ .

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About this document ...

Duane Kouba
1999-05-07