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Π / 2 ∫ 0 Sin N X Sin N X + Cos N X D X , N ∈ N .

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प्रश्न

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]
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उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x ..................(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n \left( \frac{\pi}{2} - x \right)}{\sin^n \left( \frac{\pi}{2} - x \right) + \cos^n \left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\cos^n x + \sin^n x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\sin^n x + \cos^n x} d x ................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} \right] d x \]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx\]
\[ = \int_0^\frac{\pi}{2} dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]

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अध्याय 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Very Short Answers | Q 16 | पृष्ठ ११५

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