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SOLUTIONS TO U-SUBSTITUTION



SOLUTION 19 : Integrate $ \displaystyle{ \int { e^x \cos{(e^x)} } \,dx } $ . Use u-substitution. Let

$ u = e^x $

so that

$ du = e^x dx $ .

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

$ \displaystyle{ \int { e^x \cos{(e^x)} } \,dx } = \displaystyle{ \int { \cos{(e^x)} } \,e^x dx } $

$ = \displaystyle{ \int { \cos u } \,du } $

$ = \displaystyle{ \sin u + C } $

$ = \displaystyle{ \sin {(e^x)} + C } $ .

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SOLUTION 20 : Integrate $ \displaystyle{ \int {x \sin{3x} } \,dx } $ . Use integration by parts. Let

$ u = x \ \ $ and $ \ \ dv = \sin{3x} \ dx $

so that

$ du = dx \ \ $ and $ \ \ v = (-1/3) \cos{3x} $ .

Therefore,

$ \displaystyle{ \int { x \sin{3x} } \,dx } = \displaystyle{ x (-1/3)\cos{3x} - \int { (-1/3){\cos {3x}} } \, dx } $

$ = \displaystyle{ - {x \over 3}\cos{3x} + (1/3)\int { \cos {3x} } \, dx } $

$ = \displaystyle{ - {x \over 3}\cos{3x} + (1/3) {\sin {3x} \over 3} + C} $

$ = \displaystyle{ - {x \over 3}\cos{3x} + {1\over 9}\sin {3x} + C } $ .

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SOLUTION 21 : Integrate $ \displaystyle{ \int { x^2 \cos x } \,dx } $ . Use integration by parts. Let

$ u = x^2 \ \ $ and $ \ \ dv = \cos x \ dx $

so that

$ du = 2x \ dx \ \ $ and $ \ \ v = \sin x $ .

Therefore,

$ \displaystyle{ \int { x^2 \cos x} \,dx } = \displaystyle{ x^2 \sin x - \int { 2x \sin x } \,dx } $

$ = \displaystyle{ x^2 \sin x - 2 \int { x \sin x } \,dx } $ .

Use integration by parts again. Let

$ u = x \ \ $ and $ \ \ dv = \sin x \ dx $

so that

$ du = dx \ \ $ and $ \ \ v = -\cos x $ .

Hence,

$ \displaystyle{\int { x^2 \cos x} \,dx} = \displaystyle{ x^2 \sin x - 2\int { x \sin x } \,dx }$

$ = \displaystyle{ x^2 \sin x - 2\Big\{- x\cos x - \int {(-\cos x) } \,dx \Big\} }$

$ = \displaystyle{ x^2 \sin x - 2\Big\{- x\cos x + \int { \cos x } \,dx \Big\} }$

$ = \displaystyle{ x^2 \sin x - 2\Big\{- x\cos x + \sin x \Big\} + C } $

$ = \displaystyle{ x^2 \sin x + 2 x\cos x - 2 \sin x + C } $ .

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SOLUTION 22 : Integrate $ \displaystyle{ \int { \sin{x} \cos{x} \, e^{\sin x} } \,dx } $ . Use u-substitution. Let

$ u = \sin x $

so that

$ du = \cos x \ dx $ .

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

$ \displaystyle{ \int { \sin{x} \cos{x} \, e^{\sin x} } \,dx } = \displaystyle{ \int { \sin{x} \, e^{\sin x} } \, \cos{x} \ dx } $

$ = \displaystyle{ \int { u e^u} \, du } $ .

Now use integration by parts. Let

$ w = u \ \ $ and $ \ \ dv = e^u du $

so that

$ dw = du \ \ $ and $ \ \ v = e^u $ .

Hence,

$ \displaystyle{ \int {\sin{x} \cos{x} \, e^{\sin x} } \, dx } = \displaystyle{ \int { u e^u } \,du } $

$ = \displaystyle{ u e^u - \int {e^u } \, du } $

$ = \displaystyle{ u e^u - e^u + C } $

$ = \displaystyle{ \sin x \, e^{\sin x} - e^{\sin x} + C } $ .

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SOLUTION 23 : Integrate $ \displaystyle{ \int { e^x \sin x } \,dx } $ . Use integration by parts. Let

$ u = e^x $ and $ \ \ dv = \sin x \ dx $

so that

$ du = e^x dx $ and $ \ \ v = -\cos x $ .

Therefore,

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

$ = \displaystyle{ -e^x \cos x + \int{ e^x \cos x } \, dx } $ .

Use integration by parts again. let

$ u = e^x $ and $ \ \ dv = \cos x \ dx $

so that

$ du = e^x dx $ and $ \ \ v = \sin x $ .

Hence,

$ \displaystyle{ \int{e^x \sin x} \,dx } = \displaystyle{ -e^x \cos x + \int{ e^x \cos x } \, dx } $

$ = \displaystyle{ -e^x \cos x + \Big\{ e^x \sin x - \int{ e^x \sin x } \, dx \Big\} } $

$ = \displaystyle{ -e^x \cos x + e^x \sin x - \int{ e^x \sin x } \, dx } $ .

To both sides of this "equation" add $ \displaystyle{ \int { e^x \sin x } \,dx } $ , getting

$ 2 \displaystyle{ \int { e^x \sin x } \,dx } = \displaystyle{ -e^x \cos x + e^x \sin x } + C $ .

Thus,

$\displaystyle{ \int { e^x \sin x } \,dx } = (1/2)\displaystyle{( -e^x \cos x + e^x \sin x + C ) } $

(Combine constant $ 1/2 $ with $ C $ since $ C $ is an arbitrary constant.)

$ = (1/2)\displaystyle{( -e^x \cos x + e^x \sin x) + C } $ .

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SOLUTION 24 : Integrate $ \displaystyle{ \int{\sin{3x} \cos{4x} } \,dx } $ . Use integration by parts. Let

$ u = \sin{3x} \ \ $ and $ \ \ dv = \cos{4x} \ dx $

so that

$ du = 3 \cos{3x} \ dx \ \ $ and $ \ \ v = (1/4) \sin{4x} $ .

Therefore,

$ \displaystyle{ \int{\sin{3x} \cos{4x} } \,dx } = \displaystyle{ \sin{3x} \ (1/4) \sin{4x} - \int (1/4) \sin{4x} \ (3) \cos{3x} \,dx } $

$ = \displaystyle{ (1/4) \sin{3x} \sin{4x} - (3/4) \int \cos{3x} \sin{4x} \,dx } $ .

Use integration by parts again. let

$u = \cos{3x} $ and $ \ \ dv = \sin{4x} \ dx $

so that

$ du = -3 \sin{3x} \ dx $ and $ \ \ v = (-1/4) \cos{4x} $ .

Hence,

$ \displaystyle{ \int{\sin{3x} \cos{4x} } \,dx } = \displaystyle{ (1/4) \sin{3x} \sin{4x} - (3/4) \int \cos{3x} \sin{4x} \,dx } $

$ = \displaystyle{ (1/4) \sin{3x} \sin{4x} - (3/4) \Big\{ \cos{3x} \ (-1/4) \cos{4x} - \int{ (-1/4) \cos{4x} \ (-3) \sin{3x} } \, dx \Big\} } $

$ = \displaystyle{ (1/4) \sin{3x} \sin{4x} - (3/4) \Big\{ (-1/4) \cos{3x} \cos{4x} - (3/4) \int{ \sin{3x} \cos{4x} } \, dx \Big\} } $

$ = \displaystyle{ (1/4) \sin{3x} \sin{4x} + (3/16) \cos{3x} \cos{4x} + (9/16) \int{ \sin{3x} \cos{4x} } \, dx } $ .

From both sides of this "equation" subtract $ (9/16) \int{ \sin{3x} \cos{4x} } \, dx $ , getting

$ (7/16) \displaystyle{ \int{\sin{3x} \cos{4x} } \,dx } = \displaystyle{ (1/4) \sin{3x} \sin{4x} + (3/16) \cos{3x} \cos{4x} } + C $ .

Thus,

$ \displaystyle{ \int{\sin{3x} \, \cos{4x} } \,dx } = (16/7) \Big( \displaystyle{ (1/4) \sin{3x} \sin{4x} + (3/16) \cos{3x} \cos{4x} } + C \Big) $

(Combine constant $ 16/7 $ with $ C $ since $ C $ is an arbitrary constant.)

$ = (4/7) \sin{3x} \sin{4x} + (3/7) \cos{3x} \cos{4x} + C $ .

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SOLUTION 25 : Integrate $ \displaystyle{ \int{ \sec x \ \sqrt{ \sec x + \tan x}} \,dx } $ . Use u-substitution. Let

$ u = \sqrt{ \sec x + \tan x } $

so that

$ du = (1/2)(\sec x + \tan x)^{-1/2} (\sec x \tan x + \sec^2 x)dx $

$ = (1/2)(\sec x + \tan x)^{-1/2} \sec x (\sec x + \tan x) dx $

$ = (1/2) \sec x (\sec x + \tan x)^{-1/2} (\sec x + \tan x)^1 dx $

$ = (1/2)\sec x \,(\sec x + \tan x)^{1 - 1/2} dx $

$ = (1/2)\sec x \,(\sec x + \tan x)^{1/2} dx $

$ = (1/2) \sec x \, \sqrt{ \sec x + \tan x } \ dx $ ,

or

$ 2 \ du = \sec x \, \sqrt{ \sec x + \tan x } \ dx $ .

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

$ \displaystyle{\int{\sec x \ \sqrt{ \sec x + \tan x} } \,dx } = \displaystyle{ \int{ 2 }\, du} $

$ = \displaystyle{ 2u + C } $

$ = \displaystyle{ {2\sqrt{ \sec x + \tan x}} + C} $ .

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SOLUTION 26 : Integrate $ \displaystyle{ \int{\sin{2x} - \cos{2x} \over \sin {2x} + \cos{2x}} \,dx } $ . Use u-substitution. Let

$u = \sin {2x} + \cos{2x} $

so that

$ du = (2\cos{2x} - 2\sin{2x})dx $

$ = 2(\cos{2x} - \sin{2x}) dx $

$ = -2(\sin{2x} - \cos{2x}) dx $ ,

or

$ (-1/2)du = (\sin{2x} - \cos{2x}) dx $ .

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

$ \displaystyle{\int{\sin{2x} - \cos{2x} \over \sin {2x} + \cos{2x}} \,dx } = \displaystyle{\int{ 1 \over \sin {2x} + \cos{2x}} \, ( \sin{2x} - \cos{2x} )dx } $

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

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

$ = \displaystyle{ (-1/2) \ln{\vert u\vert} + C }$

$ = \displaystyle{ (-1/2) \ln{\vert\sin {2x} + \cos{2x}\vert} + C}$ .

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SOLUTION 27 : Integrate $ \displaystyle{ \int{\sin{x} + \cos{x} \over e^{-x} + \sin {x} } \,dx } $ . First multiply by $ \displaystyle{e^x \over e^x } $ , getting

$\displaystyle{ \int{\sin{x} + \cos{x} \over e^{-x} + \sin {x} } \,dx} = \disp... ...x} + \cos{x} \over e^{-x} + \sin {x}}\Big)\, \Big({e^x \over e^x}\Big) } \,dx} $

$ = \displaystyle{ \int{{e^x \sin{x} + e^x \cos{x} \over e^{-x}e^{x} + e^x\sin {x}} } \,dx} $ .

$ = \displaystyle{ \int{{e^x \sin{x} + e^x \cos{x} \over e^{-x+x} + e^x\sin {x}} } \,dx} $ .

$ = \displaystyle{ \int{{e^x \sin{x} + e^x \cos{x} \over 1+ e^x\sin {x}} } \,dx} $ .

Now use u-substitution. Let

$ u = 1+ e^x\sin {x} $

so that

$ du = (e^x\sin{x} + e^x\cos{x})dx $ .

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

$ \displaystyle{ \int{{e^x \sin{x} + e^x \cos{x} \over 1+ e^x\sin {x}} } \,dx} ... ...playstyle{ \int{ 1 \over 1+ e^x\sin {x}} } \,( {e^x \sin{x} + e^x \cos{x} )dx} $

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

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

$ = \displaystyle{ \ln{\vert 1+ e^x\sin {x}\vert} + C}$ .

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Duane Kouba 2000-04-19