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 $\int e^{x}(x+\sin(e^{x}))~dx=\int e^{x}x~dx+\int e^{x}\sin(e^{x})~dx$.
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let $u=x$ and $dv=e^{x}dx$. Then, $du=dx$ and $v=e^{x}$. So, we have
$\int e^{x}(x+\sin(e^{x}))~dx={\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx=xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx$

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

(b)

Step 1:

Step 2:

Step 3:

(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)
(c) $-\cos x+{\frac {\cos ^{3}x}{3}}+C$

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