009B Sample Midterm 3, Problem 4

Evaluate the integral:

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

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

Solution:

Step 1:
We proceed using integration by parts.
Let $u=\sin(\ln x)$ and $dv=dx.$ Then, $du=\cos(\ln x){\frac {1}{x}}dx$ 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
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 (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.}

Step 2:
Now, we need to use integration by parts again.
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(\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=-\sin(\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.}
Therfore, 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 \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 (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 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 (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(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.}
Thus, 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 \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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