\(\int_{0}^{\pi/2}\displaystyle\frac{\sin^8 x}{\sin^8 x + \cos^8 x}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4}\)
Let
\(\displaystyle I = \int_{0}^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x}\,dx\)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\), replace x by \(\frac{\pi}{2} - x\). Since \(\sin\left(\frac{\pi}{2} - x\right) = \cos x\) and \(\cos\left(\frac{\pi}{2} - x\right) = \sin x\),
\(\displaystyle I = \int_{0}^{\pi/2} \frac{\cos^8 x}{\cos^8 x + \sin^8 x}\,dx\)
Adding the two forms of I, the numerators add up to the denominator:
\(\displaystyle 2I = \int_{0}^{\pi/2} 1\,dx = \frac{\pi}{2}\)
\(\displaystyle I = \frac{\pi}{4}\)