Question

The value of integral \(∫_0^{\frac \pi4} \frac {xdx}{cos^42x+sin^42x}\).

Answer 1

Evaluate the integral \(\displaystyle I=\int_{0}^{\pi/4}\frac{x}{\cos^{4}(2x)+\sin^{4}(2x)}\,dx\).

Let \(f(x)=\dfrac{1}{\cos^{4}(2x)+\sin^{4}(2x)}\). Using the complementary definite integral (CD2) property,

\(\displaystyle \int_0^a x f(x)\,dx=\int_0^a (a-x)f(a-x)\,dx\).

Here \(a=\frac{\pi}{4}\). Now, \(f(\frac{\pi}{4}-x)=f(x)\) since \(\cos(\frac{\pi}{2}-2x)=\sin2x\).

Thus

\(\displaystyle I=\int_0^{\pi/4}x f(x)\,dx=\int_0^{\pi/4}(\frac{\pi}{4}-x)f(x)\,dx\).

Adding them,

\(\displaystyle 2I=\frac{\pi}{4}\int_0^{\pi/4}\frac{dx}{\cos^4(2x)+\sin^4(2x)}\).

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

Using the identity

\(\cos^4\theta+\sin^4\theta=\frac{3+\cos4\theta}{4}\),

\(\displaystyle \int_0^{\pi/4}\frac{dx}{\cos^4(2x)+\sin^4(2x)}=\frac{\pi}{2\sqrt2}\).

Therefore

\(\displaystyle I=\frac{\pi}{8}\times\frac{\pi}{2\sqrt2}=\frac{\pi^2}{16\sqrt2}\).

The answer is \(\frac{\pi^2}{16\sqrt2}\).