Leicester: Integration 3

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Bill Foster

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Bill Foster 55 years, 2 months ago

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Question statement

Find the following indefinite integrals using integration by parts.

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
c	integer	5
b	integer	-5
d	integer	8
s2	integer	1
s1	integer	-1
u	integer	1
c1	integer	5
d1	integer	6

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Parts

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Part c) Mathematical expression

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

$I=\;$ Unnamed gap

Input all numbers as fractions or integers and not decimals.

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The formula for integrating by parts is

$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$

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,

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

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:

$\begin{eqnarray*}I&=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) +({2*a}/{c})*((({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d}))+C}\\ &=& \simplify[std]{-(({a}*x+{b})^2/{c})*cos({c}*x+{d})+(({2*a}({a}x+{b}))/{c^2})*sin({c}*x+{d})+({2*a^2}/{c^3})*cos({c}*x+{d})+C} \end{eqnarray*}$

Let $\displaystyle A= \simplify[std]{int(exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x)),x)}$ . We solve this using two integration by parts, and we choose $u = \simplify[std]{ {u}*sin({d1}x)+{1-u}*cos({d1}x)}$ in both.

$\begin{eqnarray*} A&=&\simplify[std]{ 1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}int(exp({c1}x) *({u}*cos({d1}x)+{(1-u)}*sin({d1}x)),x)}\\&=&\simplify[std]{1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}*(1/{c1}exp({c1}x)*( {u}*cos({d1}x)+{1-u}*sin({d1}x))+{(-1)^(u+1)*d1}/{c1}int(exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x)),x) )}\\&=&\simplify[std]{1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}*(1/{c1}exp({c1}x)*( {u}*cos({d1}x)+{1-u}*sin({d1}x))+{(-1)^(u+1)*d1}/{c1}A )}\end{eqnarray*}$

Note that after integrating by parts twice, we have the integral $A$ on both sides of this equation.

Rearranging we have:

$A = \simplify[std]{e^({c1}x)/{c1^2+d1^2}*(({u*(c1-d1)+d1})*sin({d1}x)+({u*(-c1-d1)+c1})*cos({d1}x))+C}$

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Bill Foster 55 years, 2 months ago

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Leicester: Integration 3

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Bill Foster 55 years, 2 months ago

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