Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$

Question

I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$ I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.

Do you have any ideas? :)

EDIT: Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to

$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$

By expanding with 1: $$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$ $$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$ $$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$

And using the substitution: $t=\tan\left(x\right)$

$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$

$$t^2=\tan^2\left(x\right)$$

$$t^2=\frac{\sin^2x}{\cos^2x}$$

$$t^2=\frac{1-\cos^2x}{\cos^2x}$$

$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$ $$t^2+1=\frac{1}{\cos^2x}$$

Using it: $$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$

I don't think I got to the expected result but I can't seem to be able to find why…

Answer 1

Following your substitution $t= \tan x$, integrate the resulting as follows

\begin{align} &\int \frac1{1+\sin^4x}dx\\ =&\int \:\frac{t^2+1}{2t^4+2t^2+1}dt =\int \frac{1+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\ =&\frac{\sqrt2+1}{2\sqrt2} \int \frac{\sqrt2+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt -\frac{\sqrt2-1}{2\sqrt2} \int \frac{\sqrt2-\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\ =&\frac{\sqrt2+1}{2\sqrt2} \int \frac{d(\sqrt2t-\frac1{t})}{(\sqrt2t-\frac1t)^2 + 2(\sqrt2+1)}dt -\frac{\sqrt2-1}{2\sqrt2} \int \frac{d(\sqrt2t+\frac1{t})}{(\sqrt2t+\frac1t)^2 -2(\sqrt2-1)} dt\\ =&\frac{\sqrt{\sqrt2+1}}4\tan^{-1} \frac{t-\frac1{\sqrt2 t}}{\sqrt{\sqrt2+1}} +\frac{\sqrt{\sqrt2-1}}4\coth^{-1} \frac{t+\frac1{\sqrt2 t}}{\sqrt{\sqrt2-1}}+C \end{align}

Answer 2

Double-check my arithmetic, but here's the strategy.

@Bernard's suggested substitution $t=\tan x$ gives \begin{align} & \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\ = {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align} with $c:=\sqrt{2\sqrt{2}-1}$. You can do the rest with partial fractions.

Answer 3

For $n=4$, you can make it shorter since $$A=\frac 1{1+\sin^4(x)}=\frac 1{(\sin^2(x)+i)(\sin^2(x)-i)}$$

Using partial fractions and double angle formula $$A=\frac{i}{\cos (2 x)-(1-2 i)}-\frac{i}{\cos (2 x)-(1+2 i)}$$ Using now the tangent half-angle $$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan ^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$

Answer 4

Letting $t=\tan x$ transforms the integral into $$ \begin{aligned} & I=\int \frac{1+t^2}{2 t^4+2 t^2+1} d t \\ & =\frac{1}{2} \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{2 t^2}+1} d t \\ & =\int \frac{\frac{\sqrt{2}+1}{2}\left(1+\frac{1}{\sqrt{2} t^2}\right)-\frac{\sqrt{2}-1}{2}\left(1-\frac{1}{\sqrt{2} t^2}\right)}{t^2+\frac{1}{(\sqrt{2} t)^2}+1} d t \\ & =\frac{\sqrt{2}+1}{2} \int \frac{d\left(t-\frac{1}{\sqrt{2} t}\right)}{\left(t-\frac{1}{\sqrt{2} t}\right)^2+(\sqrt{2}+1)}-\frac{\sqrt{2}-1}{2} \int \frac{d\left(t+\frac{1}{\sqrt{2} t}\right)}{\left(t+\frac{1}{\sqrt{2} t}\right)^2-(\sqrt{2}-1)} \\ & =\frac{1}{2}\left[\sqrt{\sqrt{2}+1} \tan ^{-1}\left(\frac{t-\frac{1}{\sqrt{2} t}}{\sqrt{\sqrt{2}+1}}\right)+\sqrt{\sqrt{2}-1}\tanh ^{-1}\left(\frac{t+\frac{1}{\sqrt{2} t}}{\sqrt{\sqrt{2}-1}}\right)\right]+C \\ & \end{aligned} $$