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### Integration by Substitution Exercises Worked

##### Contents

C98-9. {$$\color{blue}{ y = \int \sqrt{3x + 2} }$$}

{$$\color{red}{y={2\over 9}(3x+2)^{3 \over 2} }$$}{$$\color{green}{y={2\over 9}(3x+2)^{3 \over 2} +2}$$}{$$\color{maroon}{y= {2\over 9}(3x+2)^{3 \over 2}-2}$$}

#### C98-9. {$\displaystyle \int \sqrt{3x + 2}dx$}

Substitution:{$$u=3x +2,\; du = 3dx,\; {{du} \over 3} = dx$$} {\begin{align} & \int \sqrt{3x + 2}dx \cr &= {2\over 3}u^{3 \over 2}({1 \over 3})+ C \cr &= {2\over 9}u^{3 \over 2}+ C \cr &= {2\over 9}(3x+2)^{3 \over 2}+ C \cr \end{align}}

Check: {\begin{align} &{d \over {dx}}[ {2\over 9}(3x+2)^{3 \over 2}+ C] \cr &= {2 \over 9}({3 \over 2}) (3x+2)^{1 \over 2}(3)+0\cr &= (3x+2)^{1 \over 2} \end{align}}

C98-10. {$$\color{blue}{ y = \int {{dx} \over {3x + 2}} }$$}

{$$\color{red}{y= {1 \over 3}\ln|3x+2| }$$}{$$\color{green}{y= {1 \over 3}\ln|3x+2| +2}$$}{$$\color{maroon}{y= {1 \over 3}\ln|3x+2|-2}$$}

#### C98-10. {$\displaystyle \int {{dx} \over {3x + 2}}$}

Substitution: {$$u=3x+2,\;du=3dx,\;{{du} \over 3}=dx$$}

{\begin{align} & \int {{dx} \over {3x+2}} \cr &= \int {1 \over 3}( {{du} \over {3x +2}} ) \cr &= {1 \over 3} \int u^{-1}du \cr &= {1 \over 3}\ln |u| + C \cr &= {1 \over 3}\ln|3x + 2| + C\end{align}}

Check:

{\begin{align} &{d \over {dx}}\left( {1 \over 3}\ln|3x + 2| + C \right) \cr &= {1 \over 3}(3x+2)^{-1}(3)+ 0 \cr &= (3x+2)^{-1} \end{align}}

C98-11. {$$\color{blue}{ y = \int {{dx} \over {1+(3x + 2)^2}} }$$}

{$$\color{red}{y= {1 \over 3}\arctan(3x+2) }$$}{$$\color{green}{y= {1 \over 3}\arctan(3x+2) +2}$$}{$$\color{maroon}{y= {1 \over 3}\arctan(3x+2) -2}$$}

#### C98-11. {$\displaystyle \int {{dx} \over {1+(3x+2)^2}}$}

Substitution:

{$u=3x+2,\;du=3dx,\;{{du} \over 3}=dx$}

{\begin{align} & \int {{dx} \over {1+(3x+2)^2}} \cr &= \int {{{du} \over 3} \over {1+u^2}} \cr &= {1 \over 3}\int {1 \over {1+u^2}}du \cr &= {1 \over 3}\arctan u \cr &= {1 \over 3}\arctan (3x+2) \end{align}}

Check: {\begin{align} & {d \over {dx}}\left( {1 \over 3}\arctan (3x+2) + C \right) \cr &= {1 \over 3} \left( {1\over {1+(3x+2)^2}} \right) 3 + 0 \cr &= {1\over {1+(3x+2)^2}} \cr \end{align}}

C98-12. {$$\color{blue}{ y = x\sqrt{4+x^2} }$$}

{$$\color{red}{y= {1 \over 3}(3+x^2)^{3 \over 2}}$$}{$$\color{green}{y= {1 \over 3}(3+x^2)^{3 \over 2} +2}$$}{$$\color{maroon}{y= {1 \over 3}(3+x^2)^{3 \over 2} -2}$$}

#### C98-12. {$\displaystyle \int x\sqrt{4+x^2}dx$}

Substitution:

{$u=4+x^2,\;du=2x,\;{{du} \over {2x}}=dx$}

{\begin{align} & \int x\sqrt{4+x^2}dx \cr &= \int xu^{1\over 2}{1 \over {2x}}du \cr &= {1 \over 2}\int u^{1\over 2}du \cr &= ({1 \over 2})({2 \over 3}) u^{3\over 2} + C \cr &= {1 \over 3}(4+x^2)^{3\over 2} + C \cr \end{align}}

Check: {\begin{align} &{d \over {dx}}\left( {1 \over 3}(4+x^2)^{3\over 2} +C \right) \cr &= ({1 \over 3})({3 \over 2} )(4+x^2)^{1\over 2}2x +0 \cr &= x(4+x^2)^{1\over 2} \cr &= x\sqrt{4+x^2} \cr \end{align}}

C98-13. {$$\color{blue}{ y = {x\over {4+x^2}} }$$}

{$$\color{red}{y= \ln\sqrt{4+x^2} }$$}{$$\color{green}{y= \ln\sqrt{4+x^2} +2}$$}{$$\color{maroon}{y= \ln\sqrt{4+x^2} -2}$$}

#### C98-13. {$\displaystyle \int {{xdx} \over{4+x^2}}$}

Substitution:

{$u=4+x^2,\;du=2x,\;{{du} \over {2x}}=dx$}

{\begin{align} & \int {{xdx} \over {4+x^2}} \cr &= \int x {1 \over u}{1 \over {2x}}du \cr &= {1 \over 2}\int {1 \over u}du \cr &= {1 \over 2}\ln u + C \cr &= {1 \over 2}\ln (4+x^2) + C \cr &= \ln \sqrt{4+x^2} + C \end{align}}

Check: {\begin{align} & {d \over {dx}}\left( \ln \sqrt{4+x^2} + C \right) \cr &= {1 \over 2}\ln (4+x^2) +0 \cr &= ({1 \over 2}){1 \over{4+x^2}}2x \cr &= {x \over{4+x^2}} \cr \end{align}}

C98-14. {$$\color{blue}{ y = \sin x \sqrt {\cos x} }$$}

{$$\color{red}{y= -{2 \over 3}(\cos x)^{3 \over 2} }$$}{$$\color{green}{y= -{2 \over 3}(\cos x)^{3 \over 2} +2}$$}{$$\color{maroon}{y= -{2 \over 3}(\cos x)^{3 \over 2} -2}$$}

#### C98-14. {$\displaystyle \int \sin x \sqrt {\cos x}dx$}

Substitution:

{$u=\cos x,\;du=-\sin x,\;{{du} \over {- \sin x}}=dx$}

{\begin{align} & \int sin x \sqrt {\cos x} dx \cr &= \int \sin x \sqrt u{{-1} \over {\sin x}}du \cr &= -\int \sqrt u du \cr &= -{2 \over 3}u^{3 \over 2} + C \cr &= -{2 \over 3}(\cos x)^{3 \over 2} + C \end{align}}

Check: {\begin{align} & {d \over {dx}}\left( -{2 \over 3}(\cos x)^{3 \over 2} + C \right) \cr &= -{2 \over 3}({3 \over 2}) (\cos x)^{1 \over 2} (-\sin x) \cr &= \sin x \sqrt{\cos x} \end{align}}

C98-15. {$$\color{blue}{ y = \cos x \sqrt {\sin x} }$$}

{$$\color{red}{y= {2 \over 3}(\sin x)^{3 \over 2} }$$}{$$\color{green}{y= {2 \over 3}(\sin x)^{3 \over 2} +2}$$}{$$\color{maroon}{y= {2 \over 3}(\sin x)^{3 \over 2} -2}$$}

#### C98-15. {$\displaystyle \int \cos x \sqrt {\sin x}dx$}

Substitution:

{$u=\sin x,\;du=\cos x,\;{{du} \over { \cos x}}=dx$}

{\begin{align} & \int cos x \sqrt {\sin x} dx \cr &= \int \cos x \sqrt u{1 \over {\cos x}}du \cr &= \int \sqrt u du \cr &= {2 \over 3}u^{3 \over 2} + C \cr &= {2 \over 3} (\sin x)^{3 \over 2} + C \end{align}}

Check: {\begin{align} & {d \over {dx}}\left( {2 \over 3}(\sin x)^{3 \over 2} + C \right) \cr &= {2 \over 3}({3 \over 2}) (\sin x)^{1 \over 2} (\cos x) \cr &= \cos x \sqrt{\sin x} \end{align}}

C98-16. {$$\color{blue}{ y = \cos x(1 + 4\sin x + 9\sin^2x )}$$}

{$$\color{red}{y= \sin x + 2\sin^2x + 3\sin^3x }$$}{$$\color{green}{y= \sin x + 2\sin^2x + 3\sin^3x +2}$$}{$$\color{maroon}{y= \sin x + 2\sin^2x + 3\sin^3x -2}$$}

#### C98-16. {$\displaystyle \int \cos x(1 + 4\sin x + 9\sin^2x )dx$}

Substitution:

{$u=\sin x,\;du=\cos x,\;{{du} \over { \cos x}}=dx$}

{\begin{align} & \int \cos x(1 + 4\sin x + 9\sin^2x )dx \cr &= \int \cos xdx + \int 4\cos x\sin x dx + \int 9\cos x\sin^2x dx \cr &= \int \cos x dx + \int 4\cos x u{1 \over {\cos x}}du + \int 9\cos x u^2 {1 \over {\cos x}}du \cr &= \int \cos x dx + 4\int u du + 9\int u^2 du \cr &= \sin x + 4({1 \over 2})u^2 + 9({1 \over 3}) u^3 +C \cr &= \sin x + 2\sin^2x + 3\sin^3x +C \end{align}}

Check: {\begin{align} & {d \over {dx}}\left( \sin x + 2\sin^2x + 3\sin^3x +C \right) \cr &= \cos x 2(2\cos x\sin x) + 3(3\cos x \sin^2 x)\cr &= \cos x(1 + 4\sin x + 9\sin^2 x) \end{align}}

C98-17. {$$\color{blue}{ y = \sin^3 }$$}

{$$\color{red}{y= -\cos x + {1 \over 3}\cos^3 }$$}{$$\color{green}{y= -\cos x + {1 \over 3}\cos^3 +2}$$}{$$\color{maroon}{y= -\cos x + {1 \over 3}\cos^3 -2}$$}

#### C98-17. {$\displaystyle \int \sin^3 x dx$}

{$u=\cos x,\;du= - \sin x,\;{{-du} \over { \sin x}}=dx$}

{\begin{align} &\int \sin^3 x dx = \int \sin x(1- cos^2x) dx \cr &= \int \sin x(1 - u^2){{-1} \over { \sin x}}du \cr &= \int -1 + u^2du \cr &= -\int 1du + \int u^2du \cr &= -u + {1 \over 3} u^3 \cr &= -\cos x + {1 \over 3} \cos^3x \end{align}}

Check:

{\begin{align} & {d \over {du}}(-\cos x + {1 \over 3} \cos^3x) = -(-\sin x) + 3({1 \over 3})\cos^2 x(-\sin x) \cr &= \sin x - \sin x\cos^2 x = \sin x(1-\cos^2 x) \cr &= \sin{x} \sin^2 x = \sin^3 x \end{align}}

Graphs:

Sources:

Recommended:

Category: Math Calculus Integration

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### March 16, 2018

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