\(\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\displaystyle\frac{\sin^{4}x}{\sin^{4}x+\cos^{4}x}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{8}\)
Let
\( \displaystyle I = \int_{\pi/8}^{3\pi/8}\frac{\sin^{4}x}{\sin^{4}x+\cos^{4}x}\,dx \qquad (1) \)
Here \(a+b = \frac{\pi}{8}+\frac{3\pi}{8} = \frac{\pi}{2}\). By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\), replace x by \(\frac{\pi}{2}-x\); then \(\sin x\) becomes \(\cos x\) and \(\cos x\) becomes \(\sin x\):
\( \displaystyle I = \int_{\pi/8}^{3\pi/8}\frac{\cos^{4}x}{\cos^{4}x+\sin^{4}x}\,dx \qquad (2) \)
Adding (1) and (2), the numerators add up to the common denominator:
\( \displaystyle 2I = \int_{\pi/8}^{3\pi/8} 1\,dx \)
\( \displaystyle 2I = \frac{3\pi}{8}-\frac{\pi}{8} = \frac{\pi}{4} \)
\( \displaystyle I = \frac{\pi}{8} \)