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प्रश्न

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\]

 

योग
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उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x ..............(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{\sin^n \left( \frac{\pi}{2} - x \right)}{\sin^n \left( \frac{\pi}{2} - x \right) + \cos^n \left( \frac{\pi}{2} - x \right)} dx\]

\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\cos^n x + \sin^n x} dx \]

\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\sin^n x + \cos^n x} dx ................(2)\]

\[\text{Adding (1) and (2) we get}\]

\[2I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} d x \]

\[ = \int_0^\frac{\pi}{2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx\]

\[ = \int_0^\frac{\pi}{2} 1\ dx\]

\[ = \int_0^\frac{\pi}{2} dx\]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

\[ = \frac{\pi}{2}\]

\[Hence\ I = \frac{\pi}{4}\]

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अध्याय 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.5 | Q 5 | पृष्ठ ९४

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