STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$ = 2\int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}} - 3\int_0^{\pi /2} {\frac{{dx}}{{{{(2 + \cos x)}^2}}}} } $

$ = 4\int_0^1 {\frac{{dt}}{{3 + {t^2}}} - 6\int_0^1 {\frac{{1 + {t^2}}}{{{{(3 + {t^2})}^2}}}dt} } $,           $\left[ {\,\,\tan \frac{x}{2} = t} \right]$ रखने पर

$ = - 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} + 12\int_0^1 {\frac{{dt}}{{{{(3 + {t^2})}^2}}}} } $

$ = - 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} + 12} \left[ {\frac{1}{6}.\frac{t}{{{t^2} + 3}}} \right]_0^1 + \frac{1}{6}\int_0^1 {\frac{{dt}}{{3 + {t^2}}}} $

$ = 2\left[ {\frac{t}{{{t^2} + 3}}} \right]_0^1 = \frac{1}{2}$.