Trigonometric/Logarithmic Integration

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

Give me a clue or show step by step solutions please.

Thank you very much.

Make a substitution: $u = \ln x$. Then $du = \frac{1}{x} dx$, so $dx = x du$. Then you can use the fact that $e^u = x$.

Hope this helps!

• Thank you sir! Thank you very much. I got the answer! I will try to attempt the question to the best of my ability next time before posting! Thank you! – Alexander Collins Apr 9 '13 at 15:02

Putting $\ln x=y, x=e^y, dx=e^y dy$

So, $\int \sin(\ln x)dx=\int \sin y\cdot e^y dy$

Use Integration by parts, with $e^y$ as the first term

Alternatively, using Euler's formula, $e^{iy}=\cos y+i\sin y$

$\int \sin y\cdot e^y dy$ is the imaginary part of $\int e^y\cdot e^{iy}dy$

$$\int e^y\cdot e^{iy}dy=\int e^{y(1+i)}dy=\frac{e^y(e^{iy})}{(1+i)}=\frac{(1-i)e^y(\cos y+i\sin y)}2=\frac{e^y\{(\cos y+\sin y)+i(\sin y-\cos y)\}}2$$

$$\implies \int \sin y\cdot e^y dy=\frac{e^y(\sin y-\cos y)}2$$

$$\implies \int \sin(\ln x)dx=\frac{x(\sin(\ln x)-\cos(\ln x))}2$$

• @AlexanderCollins, my pleasure. Hope I could make the idea lucid – lab bhattacharjee Apr 9 '13 at 15:02

Let our integral be $I$. Use integration by parts. Let $u=\sin(\ln x)$ and $dv=dx$. Then $u=\frac{1}{x}\cos(\ln x)$, and we can take $v=x$. Thus $$I=x\cos(\ln x)-\int \cos(\ln x)\,dx.$$ Let $J=\int \cos(\ln x)\,dx$. The same sort of calculation as the one above yields $$J=-x\cos(\ln x)+\int \sin(\ln x)\,dx.$$ Thus $$I=x\cos(\ln x)-J\qquad\text{and}\qquad J=-x\sin(\ln x)+I.$$ Solve for $I$, and don't forget the $+C$. We get $$I=\frac{x\cos(\ln x)+x\ln(\sin x)}{2} +C.$$