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प्रश्न

\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]

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उत्तर

\[\int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{1 + \cos x}} d x\]

\[Let 1 + \cos x = t, then - \sin x dx = dt\]

\[When, x \to 0, t \to 2 and x \to \frac{\pi}{2}, t \to 1\]

Therefore, the integral becomes

\[ \int_2^1 \frac{- 1}{\sqrt{t}}dt\]

\[ = \int_1^2 \frac{1}{\sqrt{t}}dt\]

\[ = 2 \left[ \sqrt{t} \right]_1^2 \]

\[ = 2\left( \sqrt{2} - 1 \right)\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 12 | पृष्ठ १२१

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