009B Sample Final 1, Problem 4

Compute the following integrals.

a) 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}

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

c) 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 \sin^3x~dx}

Foundations:
Recall:
1. We can use integration by parts 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 u~dv=uv-\int v~du} .
2. We can write the fraction 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 \frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}} for some constants 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 A,B} .
3. We have the identity 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 \sin^2(x)=1-\cos^2(x)} .

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, $du=dx$ 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
${\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int ~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}\\\end{array}}$

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
${\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int ~dx+\int {\frac {1}{x}}~dx+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx}\\\end{array}}$

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
${\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2u}}~du}\\&&\\&=&\displaystyle {x+\ln x-{\frac {3}{2}}\ln u+C}\\\end{array}}$
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
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}x~dx}&=&\displaystyle {\int -(1-u^{2})~du}\\&&\\&=&\displaystyle {-u+{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {-\cos x+{\frac {\cos ^{3}x}{3}}+C}\\\end{array}}$

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

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