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$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$

I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows $$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after that I am not able to figure out what to do. If I use double angle formula then $$\frac{\frac{\sin2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$$ again i am clueless

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  • $\begingroup$ may be you mean $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin ^2 x \cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx.$ $\endgroup$ – DXT Sep 18 '18 at 10:26
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Using identities $$\sin x+\cos x=\sqrt{2}\cos(x-\dfrac{\pi}{4})$$ $$\sin x\cos x=\cos^2(x-\dfrac{\pi}{4})-\dfrac12$$ and then substitution $\dfrac{\pi}{4}-x=u$ gives \begin{align} I &= \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx \\ &= \int_{0}^{\frac{\pi}{4}}\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}dx \\ &= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos u(3-2\cos^2u)}\ du \\ &= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos^2u(3-2\cos^2u)}\cos u\ du \end{align} now let $\sin u=w$ therefore \begin{align} I &= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} \frac{(2w^2-1)^2}{(1-w^2)(2w^2+1)}\ dw \\ &= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} -2+\frac13\frac{1}{1-w^2}+\frac83\frac{1}{2w^2+1}\ dw \\ &= \color{blue}{-\dfrac12+\dfrac{\pi}{6}+\dfrac{\sqrt{2}}{12}\operatorname{arccoth}\sqrt{2}} \end{align}

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  • $\begingroup$ yes, I'll check it, thank you. $\endgroup$ – Nosrati Sep 19 '18 at 16:24
  • $\begingroup$ Better than mine+1). $\endgroup$ – xpaul Sep 19 '18 at 17:39
  • $\begingroup$ @xpaul thank you. $\endgroup$ – Nosrati Sep 19 '18 at 17:42
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There's a small mistake in your attempt. It's $\frac{\frac{\sin^{2}2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$

Note that we can write this as $\frac{\frac{\sin^{2}2x}{4}}{\sqrt2\sin(\frac{\pi}{4}+x)\left(1-\frac{\sin2x}{2}\right)}$. Hint: Replacing $x$ with $\frac{\pi}{4}-x$ seems like a good idea here.

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Nosrati's idea is good. However his second identity is wrong. Using identities $$\sin x+\cos x=\sqrt{2}\cos(\dfrac{\pi}{4}-x), \sin x\cos x=\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]$$ and substitution $x\to\dfrac{\pi}{4}-x$ gives \begin{eqnarray} I&=&\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx\\ &=&\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}dx\\ &=&\frac14\int_{0}^{\frac{\pi}{4}}\frac{\sin^2(2x)}{\sqrt2\cos(\frac{\pi}{4}-x)(1-\frac12[\cos^2(x-\dfrac{\pi}{4})-1])}dx\\ &=&\int_{0}^{\frac{\pi}{4}}\frac{\frac14\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]^2}{\sqrt{2}\cos(\dfrac{\pi}{4}-x)(1-\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big])}dx\\ &=&\frac1{2\sqrt2}\int_{0}^{\frac{\pi}{4}}\frac{(1-2\cos^2x)^2}{\cos x(3-2\cos^2x)}dx\\ &=&\frac1{6\sqrt2}\int_{0}^{\frac{\pi}{4}}\bigg(\frac{(1-2\cos^2x)^2}{\cos x}+\frac{2\cos x(1-2\cos^2x)^2}{3-2\cos^2x}\bigg)dx. \end{eqnarray} Noting that \begin{eqnarray} \int\frac{(1-2\cos^2x)^2}{\cos x}dx&=&\int(4 \cos ^3(x)-4 \cos (x)+\sec (x))dx\\ &=&\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)+C\\ \int\frac{2\cos x(1-2\cos^2x)^2}{(3-2\cos^2x)}dx&=&2\int\frac{(2\sin^2x-1)^2}{1+2\sin^2x}d\sin x\\ &=&2\int\bigg(-3+2\sin^2x+\frac{4}{1+2\sin^2x}\bigg)d\sin x\\ &=&2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)+C\bigg). \end{eqnarray} So \begin{eqnarray} I&=&\frac1{6\sqrt2}\bigg(\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)\\ &&+2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)\bigg)\bigg)\bigg|_0^{\pi/4}\\ &=&\frac{1}{6\sqrt2}(-3\sqrt2+\pi\sqrt2+\ln(\sqrt2+1)). \end{eqnarray}

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  • $\begingroup$ Thank you for your comment on my answer. I took some hours to find the problem after I posted it yesterday, but I couldn't find that. Now I know it was $$\sin x\cos x=\cos^2(x-\dfrac{\pi}{4})-\dfrac12$$ $\endgroup$ – Nosrati Sep 19 '18 at 16:34
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Hint: Show that your integrand is equal to $$\frac{\cos (x) \cot (x)}{\cot ^3(x)+1}$$

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