009B Sample Final 1, Problem 4

Compute the following integrals.

(a)  $\int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt$

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

(c)  $\int \sin ^{3}x~dx$

Foundations:
1. Through partial fraction decomposition, we can write the fraction
${\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}$
for some constants $A,B.$
2. We have the Pythagorean identity
$\sin ^{2}(x)=1-\cos ^{2}(x).$

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
${\begin{array}{rcl}\displaystyle {\int e^{x}(x+\sin(e^{x}))~dx}&=&\displaystyle {{\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx}.\\\end{array}}$

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,  $du=2\,dx$  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)    $x+\ln x-{\frac {3}{2}}\ln(2x+1)+C$
(c)    $-\cos x+{\frac {\cos ^{3}x}{3}}+C$

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