\({\int }_{0}^{\pi /4}\frac{{\cos }^{2}x{\sin }^{2}x}{{\left({\cos }^{3}x+{\sin }^{3}x\right)}^{2}}dx\) is equal to [JEE…

Given integral is \(I=\int_0^{\pi/4}\frac{\cos^2x\sin^2x}{(\cos^3x+\sin^3x)^2}\,dx\)

Put \(\tan x=t\)

Then \(\sec^2x\,dx=dt\), so \(dx=\frac{dt}{1+t^2}\)

Also, \(\sin x=\frac{t}{\sqrt{1+t^2}}\) and \(\cos x=\frac{1}{\sqrt{1+t^2}}\)

So, \(\cos^2x\sin^2x=\frac{t^2}{(1+t^2)^2}\)

Now, \(\cos^3x+\sin^3x=\frac{1}{(1+t^2)^{3/2}}+\frac{t^3}{(1+t^2)^{3/2}}\)

Therefore, \(\cos^3x+\sin^3x=\frac{1+t^3}{(1+t^2)^{3/2}}\)

So, \((\cos^3x+\sin^3x)^2=\frac{(1+t^3)^2}{(1+t^2)^3}\)

When \(x=0\), \(t=0\), and when \(x=\frac{\pi}{4}\), \(t=1\)

Therefore, \(I=\int_0^1 \frac{\frac{t^2}{(1+t^2)^2}}{\frac{(1+t^3)^2}{(1+t^2)^3}}\cdot\frac{dt}{1+t^2}\)

\(I=\int_0^1 \frac{t^2}{(1+t^3)^2}\,dt\)

Put \(1+t^3=u\)

Then \(3t^2\,dt=du\), so \(t^2\,dt=\frac{du}{3}\)

When \(t=0\), \(u=1\), and when \(t=1\), \(u=2\)

Therefore, \(I=\frac{1}{3}\int_1^2 \frac{du}{u^2}\)

\(I=\frac{1}{3}\left[-\frac{1}{u}\right]_1^2\)

\(I=\frac{1}{3}\left(-\frac{1}{2}+1\right)\)

\(I=\frac{1}{3}\cdot\frac{1}{2}\)

\(I=\frac{1}{6}\)

Hence, the value of the integral is \(\frac{1}{6}\)