Dear student , ∫sec2xtan2x+4 dxsubstitute u=tanx2du=sec2x2dxdx=2sec2xduSo we get,=∫24u2+4du=12∫1u2+1du=12tan-1uNow , substituting value of u=12tan-1(tanx2)+C Regards

Please solve Q. No.10

Dear student ,

\int \frac{s e c^{2} x}{\tan^{2} x + 4} d x s u b s t i t u t e u = \frac{\tan (x)}{2} d u = \frac{s e c^{2} x}{2} d x d x = \frac{2}{s e c^{2} x} d u S o w e g e t, = \int \frac{2}{4 u^{2} + 4} d u = \frac{1}{2} \int \frac{1}{u^{2} + 1} d u = \frac{1}{2} \tan^{- 1} (u) N o w, s u b s t i t u t i n g v a l u e o f u = \frac{1}{2} \tan^{- 1} (\frac{\tan (x)}{2}) + C

Regards

View Full Answer