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: $t=\tan\left(x\right)$

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




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

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…

  • $\begingroup$ Any ideas how you might approach it? What have you tried apart from Wolfram? $\endgroup$
    – John_dydx
    Mar 21, 2021 at 17:51
  • $\begingroup$ Use trig identities to rewrite $\sin^4 x$ in terms of $\sin^2 2x$ and then $\sin 4x$ $\endgroup$ Mar 21, 2021 at 17:53
  • 4
    $\begingroup$ Bioche's rules say you should use the substitution $t=\tan x$ to obtain the indefinite integral of a rational function in $t$. $\endgroup$
    – Bernard
    Mar 21, 2021 at 17:57
  • $\begingroup$ @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. $\endgroup$
    – Chrakai
    Mar 21, 2021 at 18:01
  • $\begingroup$ " which doesn't seem as the correct method": why do you say so ? $\endgroup$
    – user65203
    Mar 21, 2021 at 18:09

3 Answers 3


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}


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.

  • $\begingroup$ Do you please have step-by-step solution, because I am now somewhat lost. I edited the main question with additional info. $\endgroup$
    – Chrakai
    Mar 22, 2021 at 0:18

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 $$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan ^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$


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