009B Sample Midterm 3, Problem 4

Evaluate the integral:

$\int \sin(\ln x)~dx.$

Foundations:
Integration by parts tells us that $\int u~dv=uv-\int vdu.$
How could we break up $\sin(\ln x)~dx$ into $u$ and $dv?$
Notice that 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 (\ln x)} is one term. So, we need to 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=\sin (\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=dx.}

Solution:

Step 1:
We proceed using integration by parts. 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=\sin(\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=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=\cos(\ln x)\frac{1}{x}dx} 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 v=x.}
Therefore, we get
$\int \sin(\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.$

Step 2:
Now, we need to use integration by parts again. Let $u=\cos(\ln x)$ and $dv=dx.$ Then, $du=-\sin(\ln x){\frac {1}{x}}dx$ and $v=x.$
Therfore, we get
${\begin{array}{rcl}\displaystyle {\int \sin(\ln x)~dx}&=&\displaystyle {x\sin(\ln x)-{\bigg (}x\cos(\ln x)+\int \sin(\ln x)~dx{\bigg )}}\\&&\\&=&\displaystyle {x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx.}\\\end{array}}$

Step 3:
Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
So, if we add the integral on the right to the other side of the equation, we get
$2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x)$.
Now, we divide both sides by 2 to get
$\int \sin(\ln x)~dx={\frac {x\sin(\ln x)}{2}}-{\frac {x\cos(\ln x)}{2}}.$
Thus, the final answer is
$\int \sin(\ln x)~dx={\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))+C.$

Final Answer:
${\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))+C$

Return to Sample Exam