Evaluate $\int \frac{\arctan(\sin2x)}{\cos^2x}\ \mathrm dx, $

Question

Evaluate $$\int \frac{\arctan(\sin(2x))}{\cos^2x}dx,\ x \in (-\frac{\pi}{2},\frac{\pi}{2}) $$

Now, I have no idea how to start . I checked wolframalpha and it gives an interesting solution ( I am not sure how it arrived there ) . Any hints will be appreciated !

Answer 1

Let tanx=t

$I= \int tan^{-1}\frac{2t}{1+t^2}dt$

using by parts $I= t.tan^{-1}\frac{2t}{1+t^2}-\int\frac{2(1-t^2)}{(1+t^2)^2+4t^2}dt$

$I= t.tan^{-1}\frac{2t}{1+t^2}-\int\frac{2(\frac{1}{t^2}-1)}{(t+\frac{1}{t})^2+4}dt$

Now let $(t+\frac{1}{t})=z$

Answer 2

$\int_{-\pi/2}^{\pi/2}\frac{\arctan\sin 2x}{\cos^2x}dx=0$ because the integrand is odd, and$$I:=\int_0^{\pi/2}\frac{\arctan\sin 2x}{\cos^2x}dx=2\int_0^{\pi/4}\frac{\arctan\sin 2x}{\cos^2x}dx$$is finite. Indeed,$$\frac{I}{2}=[\tan x\cdot\arctan\sin 2x]_0^{\pi/4}-\int_0^{\pi/4}\frac{\tan x\cdot2\cos 2x dx}{1+\sin^22x}.$$Each term is finite; the integral is finite because $|\tan x|\le1$ on that range.

Edit, because it's been clarified an indefinite integral is sought: I'm not sure of a neat way to obtain this antiderivative (I'll have a think), but you can verify it by differeniation.

Answer 3

$$I=\int \frac{\arctan(\sin(2x))}{\cos^2x}dx$$ Let $u=\tan(x)$ $$I=\int\tan ^{-1}\left(\frac{2 u}{u^2+1}\right)\,du$$ Integrate by parts $$I=u\arctan\left(\dfrac{2u}{u^2+1}\right)+2\int\frac{ u \left(u^2-1\right)}{u^4+6 u^2+1}\,du$$ $$J=\int\frac{ u \left(u^2-1\right)}{u^4+6 u^2+1}\,du$$ $$\frac{ u \left(u^2-1\right)}{u^4+6 u^2+1}=\frac{u(u^2-1)}{\left(u^2+a\right) \left(u^2+b\right)}=\frac 1{a-b}\left(\frac{(a+1) u}{u^2+a}-\frac{(b+1) u}{u^2+b} \right)$$

$$J=\frac 1{2(a-b)}\left((a+1) \log \left(u^2+a\right)-(b+1) \log \left(u^2+b\right)\right)$$ Just finish if you want to go back to $x$ and have a nasty form.