009B Sample Midterm 3, Problem 4

Evaluate the integral:

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

Foundations:
1. Integration by parts tells us that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int u~dv=uv-\int vdu.}
2. How could we break up Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin(\ln x)~dx} into $u$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv?}
Notice that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin(\ln x)} is one term. So, we need to let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\sin(\ln x)} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=dx.}

Solution:

Step 1:
We proceed using integration by parts.
Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\sin(\ln x)} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=dx.} Then, $du=\cos(\ln x){\frac {1}{x}}dx$ and $v=x.$
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=dx.} Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x).}
Now, we divide both sides by 2 to get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sin(\ln x)~dx={\frac {x\sin(\ln x)}{2}}-{\frac {x\cos(\ln x)}{2}}.}
Thus, the final answer is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sin(\ln x)~dx={\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))+C.}

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

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