Question

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

Answer 1

We have,

\[I = \int_0^\frac{\pi}{2} \frac{1}{2\cos x + 4\sin x} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{2 - 2 \tan^2 \frac{x}{2} + 8\tan\frac{x}{2}} d x\]

\[\text{Putting }\tan\frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2}se c^2 \frac{x}{2}dx = dt\]

\[\text{When }x \to 0; t \to 0\]

\[\text{and }x \to \frac{\pi}{2}; t \to 1\]

\[ \therefore I = 2 \int_0^1 \frac{dt}{2 - 2 t^2 + 8t}\]

\[ = - \frac{2}{2} \int_0^1 \frac{dt}{t^2 - 4 t - 1}\]

\[ = - \int_0^1 \frac{dt}{\left( t - 2 \right)^2 - 5}\]

\[ = \int_0^1 \frac{dt}{\left( \sqrt{5} \right)^2 - \left( t - 2 \right)^2}\]

\[ = \frac{1}{2\sqrt{5}} \left[ \log\left| \frac{\sqrt{5} + t - 2}{\sqrt{5} - t + 2} \right| \right]_0^1 \]

\[ = \frac{1}{2\sqrt{5}}\left[ \log\frac{\sqrt{5} - 1}{\sqrt{5} + 1} - \log\frac{\sqrt{5} - 2}{\sqrt{5} + 2} \right] \]

\[ = \frac{1}{2\sqrt{5}}\log\left[ \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \times \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \right]\]

\[ = \frac{1}{2\sqrt{5}}\log\left[ \frac{5 + 2\sqrt{5} - \sqrt{5} - 2}{5 - 2\sqrt{5} + \sqrt{5} - 2} \right]\]

\[ = \frac{1}{2\sqrt{5}}\log\left[ \frac{\sqrt{5} + 3}{- \sqrt{5} + 3} \right]\]

\[I = \frac{1}{2\sqrt{5}}\log \left( \frac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} \right) \]

\[I = \frac{1}{2\sqrt{5}}log \left( \frac{3 + \sqrt{5}}{2} \right)^2 \]

\[I = \frac{2}{2\sqrt{5}}log \left( \frac{3 + \sqrt{5}}{2} \right) \]

\[I = \frac{1}{\sqrt{5}}log \left( \frac{3 + \sqrt{5}}{2} \right)\]