2
$\begingroup$

I've been told to evaluate the indefinite integral of this function:

$$\int \sin {\ln {x}} dx$$

I'm supposed to make a $u$-substitution in the beginning, then complete it using integration by parts. Every time I try, I just end up going in circles. Could someone please help me? Thanks!

$\endgroup$
2
$\begingroup$

Hint: Let $u = \ln{x}$. Then $du = \frac{1}{x} dx$, or $x du = dx$. Using the fact that $x = e^u$, we can write

$$\int \sin \ln{x} dx = \int e^u \sin{u}du$$

This is a common integral that can be done by using parts twice, or by recognizing that

$$\sin{u} = \operatorname{Im} e^{iu}$$

$\endgroup$
0
$\begingroup$

This integral can be done directly by using parts twice,

$u=\sin(\ln x)\Rightarrow du=\dfrac{\cos(\ln x)}{x}\,dx$ and $dv=dx \Rightarrow v=x$

$\therefore\;\;\int\sin(\ln x)\, dx = x\sin(\ln x)- \int\cos(\ln x)\, dx$

$u= \cos(\ln x)\Rightarrow du= -\dfrac{\sin(\ln x)}{x}\,dx$ and $dv=dx \Rightarrow v=x$

$\therefore\;\; \int\sin(\ln x)\, dx = x\sin(\ln x)- \left[x\cos(\ln x)+\int\sin(\ln x)\, dx\right]$

We now add $\int\sin(\ln x)\, dx$ to the left and clear,

$\therefore\;\; \int\sin(\ln x)\, dx = \dfrac{x\sin(\ln x)- x\cos(\ln x)}{2}+C$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.