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प्रश्न
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x ...............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n \left( \frac{\pi}{2} - x \right)}{\sin^n \left( \frac{\pi}{2} - x \right) + \cos^n \left( \frac{\pi}{2} - x \right)} dx ..........................\left[\text{Using }\ \int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx\right]\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\cos^n x + \sin^n x} dx \]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\sin^n x + \cos^n x} dx ...............(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx\]
\[ = \int_0^\frac{\pi}{2} dx \]
\[ = \left[ x \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]
\[i . e . , \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x = \frac{\pi}{4}\]
\[ \therefore \int_0^\frac{\pi}{2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} d x = \frac{\pi}{4}\]
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