समाकल \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\) का मान क्या है?

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NIMCET 2013 Official Paper

Answer 1

अवधारणा:

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

गणना:

मान लीजिए कि, I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\) ....(1)

I = \(\rm \displaystyle\int_{0}^{\pi/2} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+ \sqrt{\cos (\dfrac{\pi}{2}-x)}}dx\)

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

(1) और (2) जोड़कर हमारे पास है

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

2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)

2I = \(\rm[x]^\frac{\pi}{2}_0\)

I = \(\dfrac{\pi}{4}\)