Calculus – Finding the Integral of arctan(sin x) from 0 to ?/2

Question

$$\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$$

I try to solve it, but failed. Who can help me to find it?

I encountered this integral when trying to solve $\displaystyle{\int_0^\pi\frac{x\cos(x)}{1+\sin^2(x)}\,dx}$.

Answer 1

Actually I didn't post my answer as I was unable to continue it where I left but after Pranav provided link I'm posting it

$$\sin x=u\iff \frac{\,\mathrm du}{\sqrt{1-u^2}}=\,\mathrm dx$$

$$\int_0^{{\pi/2}}\arctan\left(\sin x\right)\,\mathrm dx=\int_0^{1}\frac{\arctan(u)}{\sqrt{1-u^2}}\,\mathrm du$$

Now consider parametric integral $$I(\alpha)=\int_0^{1}\frac{\arctan(\alpha t)}{\sqrt{1-t^2}}\,\mathrm dt$$

Then $$\begin{align} I'(\alpha)&= \int_{0}^{1} \frac{t}{(1+\alpha^{2}t^{2})\sqrt{1-t^{2}}} \,\mathrm dt\\ &= \frac{1}{\alpha \sqrt{1+\alpha^{2}}} \text{artanh} \left[ \frac{\alpha}{\sqrt{1+\alpha^{2}}} \right]\\ &= \frac{1}{\alpha \sqrt{1+\alpha^{2}}} \text{arsinh}(\alpha) .\end{align}$$ Now for further process see this