0-dx-3-2sinx-cosx-

Question Number 17034 by arnabpapu550@gmail.com last updated on 30/Jun/17

\int_{0}^{Π} \frac{dx}{3 + 2 sinx + cosx}

Answered by Arnab Maiti last updated on 03/Jul/17

= \int_{0}^{Π} \frac{dx}{3 + 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}

= \int_{0}^{Π} \frac{\sec^{2} \frac{x}{2} dx}{3 (1 + \tan^{2} \frac{x}{2}) + 4 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}}

put \tan \frac{x}{2} = z \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dz

= \int_{0}^{\infty} \frac{2 dz}{{2 z}^{2} + 4 z + 4}

= \int_{0}^{\infty} \frac{dz}{z^{2} + 2 z + 2} = \int_{0}^{\infty} \frac{dz}{z^{2} + 2 \times 1 \times z + 1^{2} + 1}

= \int_{0}^{\infty} \frac{dz}{{(z + 1)}^{2} + 1}

= {[\tan^{- 1} (z + 1)]}_{0}^{\infty} = [\tan^{- 1} (\infty) - \tan^{- 1} (1)]

= \frac{Π}{2} - \frac{Π}{4} = \frac{Π}{4}