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प्रश्न
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
उत्तर
\[\int_0^\frac{\pi}{2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x}{1 + \cos x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x - 1 + 1}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2 - \left( 1 + \cos x \right)}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2}{1 + \cos x}dx - \int_0^\frac{\pi}{2} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{2\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx - \int_0^\frac{\pi}{2} dx\]
\[ = 2 \int_0^\frac{\pi}{2} \frac{1 - \cos x}{\sin^2 x}dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \int_0^\frac{\pi}{2} \left( \ cosec^2 x - \ cosec x\ cotx \right) dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \left[ - cotx + \ cosec x \right]_0^\frac{\pi}{2} - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 - \frac{\pi}{2}\]
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