009B Sample Final 1, Problem 4

Compute the following integrals.

a) $\int e^{x}(x+\sin(e^{x}))~dx$

b) $\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx$

c) $\int \sin ^{3}x~dx$

Foundations:
Review $u$-substitution
Integration by parts
Partial fraction decomposition
Trig identities

Solution:

(a)

Step 1:
We first distribute to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x(x+\sin(e^x))~dx=\int e^xx~dx+\int e^x\sin(e^x)~dx} .
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=e^xdx} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=dx} and $v=e^{x}$.
So, we have
${\begin{array}{rcl}\displaystyle {\int e^{x}(x+\sin(e^{x}))~dx=}&=&\displaystyle {{\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx}\\\end{array}}$

Step 2:
Now, for the one remaining integral, we use $u$-substitution.
Let $u=e^{x}$. Then, $du=e^{x}dx$.
So, we have
${\begin{array}{rcl}\displaystyle {\int e^{x}(x+\sin(e^{x}))~dx}&=&\displaystyle {xe^{x}-e^{x}+\int \sin(u)~du}\\&&\\&=&\displaystyle {xe^{x}-e^{x}-\cos(u)+C}\\&&\\&=&\displaystyle {xe^{x}-e^{x}-\cos(e^{x})+C}\\\end{array}}$

(b)

Step 1:
First, we add and subtract $x$ from the numerator. So, we have
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx=\int ~dx+\int {\frac {1-x}{2x^{2}+x}}~dx$.

Step 2:
Now, we need to use partial fraction decomposition for the second integral.
Since $2x^{2}+x=x(2x+1)$, we let ${\frac {1-x}{2x^{2}+x}}={\frac {A}{x}}+{\frac {B}{2x+1}}$.
Multiplying both sides of the last equation by $x(2x+1)$, we get $1-x=A(2x+1)+Bx$.
If we let $x=0$, the last equation becomes $1=A$.
If we let $x=-{\frac {1}{2}}$, then we get ${\frac {3}{2}}=-{\frac {1}{2}}B$. Thus, $B=-3$.
So, in summation, we have ${\frac {1-x}{2x^{2}+x}}={\frac {1}{x}}+{\frac {-3}{2x+1}}$.

Step 3:
If we plug in the last equation from Step 2 into our final integral in Step 1, we have
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=\int ~dx+\int {\frac {1}{x}}~dx+\int {\frac {-3}{2x+1}}~dx=x+\ln x+\int {\frac {-3}{2x+1}}~dx$.

Step 4:
For the final remaining integral, we use $u$-substitution.
Let $u=2x+1$. Then, $du=2dx$ and ${\frac {du}{2}}=dx$.
Thus, our final integral becomes
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=x+\ln x+\int {\frac {-3}{2x+1}}~dx=x+\ln x+\int {\frac {-3}{2u}}~du=x+\ln x-{\frac {3}{2}}\ln u+C$.
Therefore, the final answer is
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=x+\ln x-{\frac {3}{2}}\ln(2x+1)+C$

(c)

Step 1:
First, we write $\int \sin ^{3}x~dx=\int \sin ^{2}x\sin x~dx$.
Using the identity $\sin ^{2}x+\cos ^{2}x=1$, we get $\sin ^{2}x=1-\cos ^{2}x$. If we use this identity, we have
$\int \sin ^{3}x~dx=\int (1-\cos ^{2}x)\sin x~dx$.

Step 2:
Now, we proceed by $u$-substitution. Let $u=\cos x$. Then, $du=-\sin xdx$. So we have
$\int \sin ^{3}x~dx=\int -(1-u^{2})~du=-u+{\frac {u^{3}}{3}}+C=-\cos x+{\frac {\cos ^{3}x}{3}}+C$.

Final Answer:
(a) $xe^{x}-e^{x}-\cos(e^{x})+C$
(b) $x+\ln x-{\frac {3}{2}}\ln(2x+1)+C$
(c) $-\cos x+{\frac {\cos ^{3}x}{3}}+C$

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