The value of $\displaystyle \int_{0}^{\pi/2}\sin^{4}x\,\cos^{4}x\,dx$ is –

Solution: $\sin^{4}x\cos^{4}x=\big(\sin^{2}x\cos^{2}x\big)^2 =\left(\dfrac{\sin 2x}{2}\right)^{4} =\dfrac{1}{16}\sin^{4}2x.$ Thus $J=\displaystyle\int_{0}^{\pi/2}\sin^{4}x\cos^{4}x\,dx =\dfrac{1}{16}\!\int_{0}^{\pi/2}\!\sin^{4}2x\,dx =\dfrac{1}{32}\!\int_{0}^{\pi}\!\sin^{4}u\,du.$ Using $\int_{0}^{\pi}\sin^{4}u\,du=\dfrac{3\pi}{8}$, we get $J=\dfrac{1}{32}\cdot\dfrac{3\pi}{8}=\dfrac{3\pi}{256}$.