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

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…

• Any ideas how you might approach it? What have you tried apart from Wolfram? Mar 21 at 17:51
• Use trig identities to rewrite $\sin^4 x$ in terms of $\sin^2 2x$ and then $\sin 4x$ Mar 21 at 17:53
• Bioche's rules say you should use the substitution $t=\tan x$ to obtain the indefinite integral of a rational function in $t$. Mar 21 at 17:57
• @John_dydx I have tried going by expanding with $\frac{\frac{1}{cos^4x}}{\frac{1}{cos^4x}}$. There I got to $tan^4\left(x\right)$ but didn't get further yet. Mar 21 at 18:01
• " which doesn't seem as the correct method": why do you say so ?
– user65203
Mar 21 at 18:09

@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.
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{\sqrt2t-\frac1{t}}{\sqrt{2(\sqrt2+1)}} +\frac{\sqrt{\sqrt2-1}}4\coth^{-1} \frac{\sqrt2t+\frac1{t}}{\sqrt{2(\sqrt2-1)}}+C \end{align}