I am having real trouble evaluating this indefinite integral


Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ terms.

Here is what I have already tried:

  1. Seperating denominator as $(1-\sqrt{2}\sin{x}\cos{x})(1-\sqrt{2}\sin{x}\cos{x})$ and applying partial fractions.
  2. Multiplying and dividing by $\sec^4{x}$

Please provide any help. It would be appreciated if you give a hint rather than the whole solution.

  • 1
    $\begingroup$ If I was up to it, my first try would be to use $$\sin (x)=\dfrac{2\tan (x/2)}{1+(\tan (x/2))^2}\\ \cos(x)=\dfrac{1-(\tan (x/2))^2}{1+(\tan (x/2))^2}$$ and make the substitution $u=\tan (x/2)$. However, I avoid masochism as much as possible. $\endgroup$
    – Git Gud
    Commented Sep 14, 2013 at 11:29
  • $\begingroup$ @GitGud I am scared by the thought, but I'll give it a shot anyways! Any alternative ideas? $\endgroup$
    – Cheeku
    Commented Sep 14, 2013 at 11:33
  • $\begingroup$ None other than the standard tries. I'm in hopes that there is an elegant trick that spits out the answer immediately, but since you said WA gave a complicated answer, I don't think such an easy alternative exists. $\endgroup$
    – Git Gud
    Commented Sep 14, 2013 at 11:37

2 Answers 2



If we set $\int(\sin x+\cos x)dx=\sin x-\cos x,\sin2x=1-u^2$

$$\implies \int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx=2\int\frac{du}{2-(1-u^2)^2}=2\int\frac{du}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}$$



Now apply $\displaystyle\int\frac{dy}{a^2+y^2}=\frac1a\arctan \frac ya+K$

and $\displaystyle\int\frac{dy}{a^2-y^2}=\frac1{2a}\ln\big|\frac{a+y}{a-y}\big|+C$


We first split the integral into two. $$ \begin{aligned} I:=& \int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x \\ =&-\int \frac{d(\cos x)}{\left(1-\cos ^{2} x\right)^{2}+\cos ^{4} x}+\int \frac{d(\sin x)}{\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}} \\ =& \frac{1}{2}\left[\underbrace{-\int \frac{d y}{y^{4}-y^{2}+\frac{1}{2}}}_{J}+\underbrace{\int \frac{d z}{z^{4}-z^{2}+\frac{1}{2}}}_{K}\right] \end{aligned} $$

where $y=\cos x$ and $z=\sin x$

We need to find one of them, say $J$. We are going to play a little trick on $J$.

\begin{aligned}J &=\int \frac{\frac{1}{y^{2}} d y}{y^{2}+\frac{1}{2 y^{2}}-1} \\ &=\frac{1}{\sqrt 2} \int \frac{1+\frac{1}{y^{2}}-\left(1-\frac{1}{y^{2}}\right)}{y^{2}+\frac{1}{2 y^{2}}-1} d y \\ &=\frac{1}{\sqrt2} \int \frac{d\left(y-\frac{1}{y}\right)}{\left(y-\frac{1}{y}\right)^{2}+1}-\frac{1}{\sqrt2} \int \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^{2}-3}\\ &=\frac{1}{\sqrt2} \tan ^{-1}\left(y-\frac{1}{\sqrt2y}\right)-\frac{1}{2 \sqrt{6}} \ln \left|\frac{y+\frac{1}{\sqrt2y}-\sqrt{3}}{y+\frac{1}{\sqrt2y}+\sqrt{3}}\right|+C_{1} \\ &=\frac{1}{\sqrt2} \tan ^{-1}(\cos x-\frac{\sec x}{\sqrt2})-\frac{1}{2 \sqrt{6}} \ln \left|\frac{\sqrt2\cos ^{2} x-\sqrt{6} \cos x+1}{\sqrt2\cos ^{2} x+\sqrt{6} \cos x+1}\right|+C_{1}\end{aligned} Replacing y by z yields $$ K=\frac{1}{2} \tan ^{-1}(\sin x-\frac{\csc x}{\sqrt2})-\frac{1}{2 \sqrt{6}} \ln \left|\frac{\sqrt2\sin ^{2} x-\sqrt{6} \sin x+1}{\sqrt2\sin ^{2} x+\sqrt{6} \sin x+1}\right|+C_{2} $$ We can now conclude that $$ \begin{aligned} I=& \frac{1}{2\sqrt2}\left[\tan ^{-1}(\sin x-\frac{\csc x}{\sqrt2})-\tan ^{-1}(\cos x-\frac{\sec x}{\sqrt2})\right] \\ &+\frac{1}{4 \sqrt{6}} \ln \left| \frac{\left(\sqrt2\cos ^{2} x-\sqrt{6} \cos x+1\right)\left(\sqrt2\sin ^{2} x+\sqrt{6} \sin x+1\right)}{\left(\sqrt2\cos ^{2} x+\sqrt{6} \cos x+1\right)\left(\sqrt2\sin ^{2} x-\sqrt{6} \sin x+1\right)}\right|+C \end{aligned} $$


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