Correct Answer - Option 1 :
\(\dfrac{\pi}{4}\)
Concept:
\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)
Calculation:
Consider I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(i)
⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 (π/2 -x)}}{\sqrt{\sin^8 (π/2 -x)}+ \sqrt{\cos^8 (π/2 -x)}}dx\)
⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(ii)
Add (i) and (ii), we get
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}+ \sqrt{sin^8x}}{\sqrt{\cos^8 x}+ \sqrt{\sin^8 x}}dx\)
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)
⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)
⇒ I = \(\dfrac{\pi}{4}\)
