\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) యొక్క విలువను కనుగొనండి

Question

Question

Answer 1

కాన్సెప్ట్:

\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)

సాధన:

I పరిగణించండి= \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(i)

⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)

⇒ I = -\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)---(ii)

(i) మరియు (ii) జోడించండి, మేము పొందుతాము

⇒ 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 = \(\dfrac{\pi}{4}\)