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प्रश्न
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
उत्तर
\[\int_0^\pi \frac{1}{6 - \cos x} d x\]
\[ = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{6 + 6 \tan^2 \frac{x}{2} - 1 + \tan^2 \frac{x}{2}} d x\]
\[ = \int_0^\pi \frac{se c^2 \frac{x}{2}}{5 + 7 \tan^2 \frac{x}{2}}dx\]
\[Let, \tan\frac{x}{2} = t, then \frac{1}{2}se c^2 \frac{x}{2} dx = dt\]
Therefore the integral becomes
\[ \int_0^\infty \frac{2dt}{5 + 7 t^2} \]
\[ = \frac{2}{7} \int_0^\infty \frac{dt}{\frac{5}{7} + t^2} \]
\[ = \frac{2}{\sqrt{35}} \left[ \tan^{- 1} \frac{\sqrt{7}t}{\sqrt{5}} \right]_0^\infty \]
\[ = \frac{\pi}{\sqrt{35}}\]
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