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प्रश्न
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
पर्याय
π/4
π/8
π/2
0
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उत्तर
π/4
\[\int_0^\pi \frac{1}{5 + 3 \cos x} d x\]
\[ = \int_0^\pi \frac{1}{5 + 3 \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]
\[ = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{5 + 5 \tan^2 \frac{x}{2} + 3 - 3 \tan^2 \frac{x}{2}}dx\]
\[ = \int_0^\pi \frac{se c^2 \frac{x}{2}}{8 + 2 \tan^2 \frac{x}{2}}dx\]
\[Let\ \tan\frac{x}{2} = t, \text{then }\sec^2 \frac{x}{2} dx = 2dt\]
\[When\ x = 0, t = 0, x = \pi, t = \infty \]
\[\text{Therefore the integral becomes}\]
\[\frac{1}{2} \int_0^\infty \frac{dt}{4 + t^2}\]
\[ = \frac{1}{2} \left[ \tan^{- 1} \frac{t}{2} \right]_0^\infty \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}\]
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