Integration by parts

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Question 1

Find the following indefinite integrals.

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

a)

$I=\displaystyle \int \simplify[std]{({a}x+{b})*cos({c}x+{d})} dx$

$I=\;$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\frac{ 5 x - 6 }{ 3 } \sin \left ( 3 x + 7 \right ) + \frac{ 5 }{ 9 } \cos \left ( 3 x + 7 \right ) + C}$

Input all numbers as fractions or integers and not decimals.

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The formula for integrating by parts is
$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$

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b)

Use the result from the first part to find:

$\displaystyle I=\int \simplify[std]{({a}x+{b})^2*sin({c}*x+{d})} dx$

$I=\;$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-\frac{ \left ( 5 x - 6 \right )^{ 2 } }{ 3 } \cos \left ( 3 x + 7 \right ) + \frac{ 10 \left ( 5 x - 6 \right ) }{ 9 } \sin \left ( 3 x + 7 \right ) + \frac{ 50 }{ 27 } \cos \left ( 3 x + 7 \right ) + C}$

Input all numbers as fractions or integers and not decimals.

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Advice

a)

The formula for integrating by parts is
$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$

We choose $u = \simplify[std]{({a}x+{b})}$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{cos({c}*x+{d})}$ .

So $\displaystyle \frac{du}{dx} = \simplify[std]{{a}}$ and $\displaystyle v = \simplify[std]{(1/{c})*sin({c}*x+{d})}$ .

Hence,
$\begin{eqnarray} \int \simplify[std]{({a}*x+{b})*cos({c}*x+{d})} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{(({a}*x+{b})/{c})*sin({c}*x+{d}) - ({a}/{c})*Int(sin({c}*x+{d}),x)} \\ &=& \simplify[std]{(({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d}) + C} \end{eqnarray}$

b)

For this part we choose $u = \simplify[std]{({a}x+{b})^2}$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{sin({c}*x+{d})}$ .

So $\displaystyle \frac{du}{dx}=\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\displaystyle v = \simplify[std]{-(1/{c})*cos({c}*x+{d})}$ .

Hence,
$\begin{eqnarray*}I= \int \simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) + (1/{c})*Int({2*a}*({a}x+{b})*cos({c}*x+{d}),x)} \\ &=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) +({2*a}/{c})*Int(({a}x+{b})*cos({c}*x+{d}),x)}\dots (*) \end{eqnarray*}$

But in Part a we have aready worked out $\displaystyle \simplify[std]{Int(({a}x+{b})*cos({c}*x+{d}),x)=(({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d})}$

So on substituting this in equation (*) we find:
$\begin{eqnarray*}I&=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) +({2*a}/{c})*((({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d}))+C}\\ &=& \simplify[std]{-(({a}*x+{b})^2/{c})*cos({c}*x+{d})+(({2*a}({a}x+{b}))/{c^2})*sin({c}*x+{d})+({2*a^2}/{c^3})*cos({c}*x+{d})+C} \end{eqnarray*}$

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Integration by parts

a)

b)

Advice

a)

b)