Question

${ \boxed{ \bold \red{ \int \limits^{ \frac{\pi}{2} }_{0} \frac{ \sqrt{ \sin x} }{ \sqrt{ \sin x } + \sqrt{ \cos x}} \: dx}}}$






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Answer 1

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}}\:dx \cdot\cdot\cdot\cdot(1)}$

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sin(\frac{\pi}{2}-x)}}{\sqrt{sin(\frac{\pi}{2}-x)}+\sqrt{cos(\frac{\pi}{2}-x)}}}}$

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}}\:dx \cdot \cdot \cdot \cdot(2)}$

${\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}}\:dx+\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}}\:dx}}$

${\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}+\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}]\:dx}}$

$\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}]\:dx}$

$\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[1]\:dx}$

$\bold \red{2I=[x]^{\frac{\pi}{2}}_0}$

$\bold \red{2I=\frac{\pi}{2}}$

$\bold \red{2I=\frac{\pi}{2}}$

$\bold \red{I=\frac{\pi}{4}}$

Answer 2

Answer:

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}}\:dx \cdot\cdot\cdot\cdot(1)}$

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sin(\frac{\pi}{2}-x)}}{\sqrt{sin(\frac{\pi}{2}-x)}+\sqrt{cos(\frac{\pi}{2}-x)}}}}$

$\bold \red{I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}}\:dx \cdot \cdot \cdot \cdot(2)}$

${\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}}\:dx+\int\limits^{\frac{\pi}{2}}_0{\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}}\:dx}}$

${\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}+\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}]\:dx}}$

$\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}]\:dx}$

$\bold \red{2I=\int\limits^{\frac{\pi}{2}}_0[1]\:dx}$

$\bold \red{2I=[x]^{\frac{\pi}{2}}_0}$

$\bold \red{2I=\frac{\pi}{2}}$

$\bold \red{2I=\frac{\pi}{2}}$

$\bold \red{I=\frac{\pi}{4}}$

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