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प्रश्न
उत्तर
\[\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{3}{2}} d x\]
\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{3}{2}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x\]
\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 - \cos^2 x}}{\left( 1 - \cos x \right)^2}dx\]
\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sin x}{\left( 1 - \cos x \right)^2}dx\]
\[Let\ 1 - \cos x = t, Then\ \sin x\ dx = dt\]
\[When\ x = \frac{\pi}{3}, t = \frac{1}{2} and\ x\ = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[ = \int_\frac{1}{2}^1 \frac{dt}{t^2}\]
\[ = \left[ - \frac{1}{t} \right]_\frac{1}{2}^1 \]
\[ = - 1 + 2\]
\[ = 1\]
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