Question

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \frac{1}{5 + 4 \sin x} d x . Then, \]
\[I = \int_0^\frac{\pi}{2} \frac{1}{5 + 4\left( \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{5\left( 1 + \tan^2 \frac{x}{2} \right) + 8 \tan \frac{x}{2}} dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{5 \tan^2 \frac{x}{2} + 8 \tan \frac{x}{2} + 5} dx\]
\[Let\ \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]
\[When\ x = 0, t = 0 and x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = 2 \int_0^1 \frac{1}{5 t^2 + 8t + 5} dt\]
\[ \Rightarrow I = 2 \int_0^1 \frac{1}{\left( \sqrt{5}t \right)^2 + 8t + 5 + \left( \frac{4}{\sqrt{5}} \right)^2 - \left( \frac{4}{\sqrt{5}} \right)^2} dt\]
\[ \Rightarrow I = 2 \int_0^1 \frac{1}{\left( \sqrt{5}t + \frac{4}{\sqrt{5}} \right)^2 + \frac{9}{5}} dt\]
\[ \Rightarrow I = \frac{2}{3} \left[ \tan^{- 1} \left( \frac{\sqrt{5}t + \frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}} \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{3}\left[ \tan^{- 1} 3 - \tan^{- 1} \frac{4}{3} \right]\]
\[ \Rightarrow I = \frac{2}{3}\left[ \tan^{- 1} \left( \frac{3 - \frac{4}{3}}{1 + 3 \times \frac{4}{3}} \right) \right]\]
\[ \Rightarrow I = \frac{2}{3} \tan^{- 1} \frac{1}{3}\]