# 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
$\sin ^{2}(x)+\cos ^{2}(x)=1.$ 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
${\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int 1~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}.\\\end{array}}$ 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

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

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

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