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 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, 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
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}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.}
Multiplying both sides of the last equation by  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(2x+1),}
we 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 1-x=A(2x+1)+Bx.}
If we 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 x=0,} the last equation becomes  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 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
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 \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
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\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.}

(c)

Step 1:
First, we write
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=\int \sin^2 x \sin x~dx.}
Using 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^2x+\cos^2x=1,}   we 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 \sin^2x=1-\cos^2x.}
If we use this identity, we have
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=\int (1-\cos^2x)\sin x~dx.}

Step 2:
Now, we proceed by  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} -substitution.
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=\cos x.}   Then,  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 du=-\sin x dx.}
So we have
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 \begin{array}{rcl} \displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}.\\ \end{array}}

Final Answer:
(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 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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