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Evaluate $$ \int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx $$

I used the substitution $\sin x =t$, then I got the integral as $$\int_0^1 \frac{t}{2t^4-2t^2+1}dt $$

After that I don't know how to proceed. Please help me with this.

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  • $\begingroup$ You see how the equation looks very ugly right now? The answer to this problem is typesetting it with MathJax. I would strongly recommend using it. $\endgroup$ – Matti P. Aug 2 '18 at 8:18
  • $\begingroup$ I am new here, I will start using from the next question. Thanks! $\endgroup$ – balaji Aug 2 '18 at 8:20
  • $\begingroup$ Looks like a substitution $u=t^2$ is called for. But your next question; why not edit this one using MathJax? $\endgroup$ – Lord Shark the Unknown Aug 2 '18 at 8:22
  • $\begingroup$ Thanks. I got the final answer as π /4. is it correct? @LordSharktheUnknown $\endgroup$ – balaji Aug 2 '18 at 8:28
  • $\begingroup$ PARI/GP approves the result numerically, so you apparently got it. $\endgroup$ – Peter Aug 2 '18 at 8:29
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Do some trigonometry first: \begin{align} \frac{\sin x\cos x}{\sin^4x+\cos^4x}&=\frac{\tfrac12\sin 2x}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}=\frac{\tfrac12\sin 2x}{1-\frac12\sin^22x}\\&=\frac{\sin 2x}{2-\sin^22x}=\frac{\sin 2x}{1+\cos^22x}. \end{align}

Next use substitution: set $\;u=\cos 2x$, $\;\mathrm d u=-2\sin 2x\,\mathrm d x$.

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  • $\begingroup$ nice solution :-) +1 I like it. $\endgroup$ – Math-fun Aug 2 '18 at 11:38
  • $\begingroup$ @Math-fun: Thanks for your kind appreciation. It seems great minds think together ;o) $\endgroup$ – Bernard Aug 2 '18 at 11:55
  • $\begingroup$ I enjoyed it and in fact typed it but then deleted upon seeing yours :-) $\endgroup$ – Math-fun Aug 2 '18 at 12:44
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Hint:

$$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\tan x\sec^2x}{\tan^4x+1}$$

Set $\tan^2x=y$

OR $$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\cot x\csc^2x}{\cot^4x+1}$$

Set $\cot^2x=u$

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If you want to continue your solution, then with substitution $t^2=u$ $$I=\int_0^1\dfrac{t}{2t^4-2t^2+1}dt=\dfrac12\int_0^1\dfrac{1}{2u^2-2u+1}du=\int_0^1\dfrac{1}{(2u-1)^2+1}du$$ and then with substitution $2u-1=w$ $$I=\dfrac12\int_{-1}^1\dfrac{1}{w^2+1}dw=\color{blue}{\dfrac{\pi}{4}}$$

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Letting $u=\tan x$, one has \begin{eqnarray} &&\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\ &=&\int_0^{\frac{\pi}{2}}\frac{\tan(x)\sec^2(x)}{\tan^4(x)+1}dx\\ &=&\int_0^\infty\frac{u}{u^4+1}du\\ &=&\frac12\int_0^\infty\frac{1}{u^2+1}du\\ &=&\frac12\arctan(u)\bigg|_0^\infty\\ &=&\frac\pi4. \end{eqnarray}

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Perform the change of variable $y=\sin^2 x$,

$\begin{align}J&=\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\ &=\frac{1}{2}\int_0^1 \frac{1}{x^2+(1-x)^2}\,dx\\ &=\frac{1}{2}\int_0^1 \frac{1}{2x^2-2x+1}\,dx\\ &=\int_0^1 \frac{1}{(2x-1)^2+1}\,dx\\ \end{align}$

Perform the change of variable $y=2x-1$,

$\begin{align} J&=\frac{1}{2}\int_{-1}^1 \frac{1}{x^2+1}\,dx\\ &=\frac{1}{2}\Big[\arctan x\Big]_{-1}^1\\ &=\frac{1}{2}\times\frac{\pi}{2}\\ &=\boxed{\frac{\pi}{4}} \end{align}$

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