009B Sample Midterm 3, Problem 5 Detailed Solution

Evaluate the indefinite and definite integrals.

(a)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\ln x~dx}

(b)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx}

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Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$
2. Recall the trig identity
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.}

Solution:

(a)

Step 1:
To evaluate this integral, we use integration by parts.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\ln x}   and  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 dv=x~dx.}
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=\frac{1}{x}dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v={\frac {x^{2}}{2}}.}

Step 2:
Using integration by parts, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x\ln x~dx}&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-\int {\frac {x}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-{\frac {x^{2}}{4}}+C.}\end{array}}}

(b)

Step 1:
One of the double angle formulas is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos(2x)=1-2\sin ^{2}(x).}
Solving for  $\sin ^{2}(x),$  we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.}
Plugging this identity into our integral, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}}

Step 2:
If we integrate the first integral, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\left.{\frac {x}{2}}\right|_{0}^{\pi }-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}}

Step 3:
For the remaining integral, we need to use  $u$-substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=2x.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2~dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{2}}=dx.}
Also, since this is a definite integral and we are using $u$-substitution,
we need to change the bounds of integration.
We have  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{1}=2(0)=0}   and  $u_{2}=2(\pi )=2\pi .$
So, the integral becomes
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {{\frac {\pi }{2}}-\int _{0}^{2\pi }{\frac {\cos(u)}{4}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-\left.{\frac {\sin(u)}{4}}\right|_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-{\bigg (}{\frac {\sin(2\pi )}{4}}-{\frac {\sin(0)}{4}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}.}\\\end{array}}}

Final Answer:
(a)     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {x^{2}\ln x}{2}}-{\frac {x^{2}}{4}}+C}
(b)     ${\frac {\pi }{2}}$

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