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. Recall the Pythagorean identity
$\sin ^{2}(x)=1-\cos ^{2}(x).$

Solution:

(a)

Step 1:
We first note that
$\int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt=\int {\frac {t^{2}}{\sqrt {1-(t^{3})^{2}}}}~dt.$
Now, we proceed by  $u$-substitution.
Let  $u=t^{3}.$
Then,   $du=3t^{2}dt$  and  ${\frac {du}{3}}=t^{2}dt.$
So, we have
$\int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt=\int {\frac {1}{3{\sqrt {1-u^{2}}}}}~du.$

Step 2:
Now, we need to use trig substitution.
Let  $u=\sin \theta .$  Then,  $du=\cos \theta d\theta .$
So, we have
${\begin{array}{rcl}\displaystyle {\int {\frac {t^{2}}{\sqrt {1-t^{6}}}}~dt}&=&\displaystyle {\int {\frac {\cos \theta }{3{\sqrt {1-\sin ^{2}\theta }}}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {\cos \theta }{3{\sqrt {\cos ^{2}\theta }}}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {\cos \theta }{3\cos \theta }}d\theta }\\&&\\&=&\displaystyle {\int {\frac {1}{3}}~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{3}}\theta +C}\\&&\\&=&\displaystyle {{\frac {1}{3}}\arcsin(u)+C}\\&&\\&=&\displaystyle {{\frac {1}{3}}\arcsin(t^{3})+C.}\end{array}}$

(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 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)    ${\frac {1}{3}}\arcsin(t^{3})+C$
(b)    $x+\ln x-{\frac {3}{2}}\ln(2x+1)+C$
(c)    $-\cos x+{\frac {\cos ^{3}x}{3}}+C$

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