The value of the intergral \(\int_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}\) equals: [JEE Main 2024, 1 Feb (S…

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

Let \(f(x)=\frac{1}{\sin^4(2x)+\cos^4(2x)}\)

Now, \(f\left(\frac{\pi}{4}-x\right)=\frac{1}{\sin^4\left(\frac{\pi}{2}-2x\right)+\cos^4\left(\frac{\pi}{2}-2x\right)}\)

So, \(f\left(\frac{\pi}{4}-x\right)=\frac{1}{\cos^4(2x)+\sin^4(2x)}=f(x)\)

Using the property, if \(f(a-x)=f(x)\), then \(\int_0^a xf(x)\,dx=\frac{a}{2}\int_0^a f(x)\,dx\)

Here \(a=\frac{\pi}{4}\)

Therefore, \(I=\frac{\pi}{8}\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}\)

Let \(J=\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}\)

Put \(2x=t\)

Then \(2dx=dt\), so \(dx=\frac{dt}{2}\)

When \(x=0\), \(t=0\), and when \(x=\frac{\pi}{4}\), \(t=\frac{\pi}{2}\)

So, \(J=\frac{1}{2}\int_0^{\pi/2}\frac{dt}{\sin^4t+\cos^4t}\)

Now put \(\tan t=u\)

Then \(dt=\frac{du}{1+u^2}\)

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

So, \(\sin^4t+\cos^4t=\frac{u^4+1}{(1+u^2)^2}\)

Therefore, \(\frac{dt}{\sin^4t+\cos^4t}=\frac{1+u^2}{1+u^4}\,du\)

When \(t=0\), \(u=0\), and when \(t=\frac{\pi}{2}\), \(u\to\infty\)

Thus, \(J=\frac{1}{2}\int_0^\infty \frac{1+u^2}{1+u^4}\,du\)

Now, \(u^4+1=(u^2+\sqrt2u+1)(u^2-\sqrt2u+1)\)

Also, \(\frac{1+u^2}{1+u^4}=\frac{1}{2}\left(\frac{1}{u^2+\sqrt2u+1}+\frac{1}{u^2-\sqrt2u+1}\right)\)

So, \(\int_0^\infty \frac{1+u^2}{1+u^4}\,du=\frac{\pi}{\sqrt2}\)

Therefore, \(J=\frac{1}{2}\cdot\frac{\pi}{\sqrt2}=\frac{\pi}{2\sqrt2}\)

Now, \(I=\frac{\pi}{8}\cdot J\)

\(I=\frac{\pi}{8}\cdot\frac{\pi}{2\sqrt2}\)

\(I=\frac{\pi^2}{16\sqrt2}\)

\(I=\frac{\sqrt2\pi^2}{32}\)

Hence, the value of the integral is \(\frac{\sqrt2\pi^2}{32}\)