How do we evaluate $\int{\dfrac{dx}{\sin^{4}{x}+1}}$?

Question

How do we evaluate $$\displaystyle\int{\dfrac{dx}{\sin^{4}{x}+1}}?$$ don't know how to solve this, please help.

Answer 1

$\displaystyle \int \dfrac{dx}{\sin^4 x+1}$

Multiplying top and bottom with $\csc^4 x$ gives

$\displaystyle \int \dfrac{\csc^4 x}{1+\csc^4 x}\,dx$

$=\displaystyle \int \dfrac{\csc^2 x(1+\cot^2 x)}{1+(1+\cot^2 x)^2}\,dx$

Let $u=\cot x\implies du=-\csc^2 x\,dx$

$=\displaystyle \int \dfrac{-(1+u^2)}{1+(1+u^2)^2}\,du$

$=\displaystyle -\int \dfrac{1+u^2}{(1+u^2)^2-i^2}\,du$

$=\displaystyle -\int \dfrac{1+u^2}{(u^2+1+i)(u^2+1-i)}\,du$

$=\displaystyle -\dfrac{1}{2}\int \dfrac{(u^2+1+i)+(u^2+1-i)}{(u^2+1+i)(u^2+1-i)}\,du$

$=\displaystyle -\dfrac{1}{2}\int \left[\dfrac{1}{u^2+1-i}+\dfrac{1}{u^2+1+i}\right]\,du$

$=\displaystyle -\dfrac{1}{2}\int \left[\dfrac{1}{u^2+(\sqrt{1-i})^2}+\dfrac{1}{u^2+(\sqrt{1+i})^2}\right]\,du$

$=-\dfrac{1}{2}\left[\dfrac{1}{\sqrt{1-i}}\arctan\left(\dfrac{u}{\sqrt{1-i}}\right)+\dfrac{1}{\sqrt{1+i}}\arctan\left(\dfrac{u}{\sqrt{1+i}}\right)\right]+C$

$=-\dfrac{1}{2}\left[\dfrac{1}{\sqrt{1-i}}\arctan\left(\dfrac{\cot x}{\sqrt{1-i}}\right)+\dfrac{1}{\sqrt{1+i}}\arctan\left(\dfrac{\cot x}{\sqrt{1+i}}\right)\right]+C$

Answer 2

By substituting $x=\arctan t$ the problem boils down to

$$ \int \frac{1+t^2}{(1+t^2)^2+t^4}\,dt = \int\left(\frac{1+i}{4}\cdot\frac{1}{t^2+\frac{1+i}{2}}+\frac{1-i}{4}\cdot\frac{1}{t^2+\frac{1-i}{2}}\right)\,dt $$ leading to: $$ \frac{1}{2\sqrt{2}}\left(\sqrt{1-i}\,\arctan\frac{t}{\sqrt{\frac{1-i}{2}}}+\sqrt{1+i}\,\arctan\frac{t}{\sqrt{\frac{1+i}{2}}}\right) $$ that equals $$ 2^{-7/4}\left[\sqrt{2+\sqrt{2}}\arctan\frac{t\sqrt{2+2\sqrt{2}}}{1-t^2\sqrt{2}}+\sqrt{2-\sqrt{2}}\,\text{arctanh}\frac{t\sqrt{2\sqrt{2}-2}}{1+t^2\sqrt{2}}\right]$$ by exploiting $\arctan(a)\pm\arctan(b)=\arctan\frac{a\pm b}{1\mp ab}$.