7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

Put $\sin x - \cos x = t$, then

$(\sin x + \cos x)dx = dt$

$I = \int_{ - 1}^0 {\frac{{dt}}{{9 + 16(1 - {t^2})}}} $

$= \int_{ - 1}^0 {\frac{{dt}}{{25 - 16{t^2}}}} $

$ = \frac{1}{{10}}\int_{ - 1}^0 {\left( {\frac{1}{{5 - 4t}} + \frac{1}{{5 + 4t}}} \right)dt} $

$ = \left| {\frac{1}{{10}}.\frac{1}{4}[\log (5 + 4t) - \log (5 - 4t)]\,} \right|_{ - 1}^0$

$ = \frac{1}{{40}}(\log 9 - \log 1) = \frac{1}{{20}}\log 3$.