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 $v=x.$
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:
${\frac {x}{2}}(\sin(\ln x)-\cos(\ln x))+C$

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