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प्रश्न
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उत्तर
\[Let\, I = \int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[ = \int_0^{\pi/2} \frac{x + \sin x}{2 \cos^2 \frac{x}{2}} dx\]
\[ = \int_0^{\pi/2} \left[ \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{\sin x}{2 \cos^2 \frac{x}{2}} \right]dx\]
\[ = \frac{1}{2} \int_0^{\pi/2} x se c^2 \frac{x}{2}dx + \int_0^{\pi/2} \frac{2\sin \frac{x}{2}\cos\frac{x}{2}}{2 \cos^2 \frac{x}{2}}dx\]
\[ = \frac{1}{2} \left[ x \frac{\tan\frac{x}{2}}{\frac{1}{2}} \right]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{\tan\frac{x}{2}}{\frac{1}{2}}dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ x \tan\frac{x}{2} \right]_0^{\pi/2} - \int_0^{\pi/2} \tan\frac{x}{2} dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ \frac{\pi}{2} \tan\frac{\pi}{4} \right]\]
\[ = \frac{\pi}{2} \times 1\]
\[ = \frac{\pi}{2} \]
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