Question

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \frac{1}{5 \cos x + 3 \sin x} d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \frac{1}{5\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + 3\left( \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)} d\ x \left[ \because \sin A = \left( \frac{2 \tan \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} \right), \cos A = \left( \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} \right) \right]\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{5 - 5 \tan^2 \frac{x}{2} + 6 \tan \frac{x}{2}} dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{5 - 5 \tan^2 \frac{x}{2} + 6 \tan \frac{x}{2}} dx\]
\[Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]
\[Also, x = 0, t = 0 and x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \frac{2dt}{5 - 5 t^2 + 6t}\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{2dt}{1 - t^2 + \frac{6}{5}t + \frac{36}{100} - \frac{36}{100}}\]
\[ = \frac{2}{5} \int_0^1 \frac{dt}{- \left( t - \frac{6}{10} \right)^2 + \frac{136}{100}}\]
\[ = \frac{2}{5} \times \frac{10}{\sqrt{136}} \left[ - \log \left( \frac{t - \frac{6}{10} - \frac{\sqrt{136}}{10}}{t - \frac{6}{10} + \frac{\sqrt{136}}{10}} \right) \right]_0^1 \]
\[ = \frac{1}{\sqrt{34}}\left[ - \log \left( \frac{4 - 2\sqrt{34}}{4 + 2\sqrt{34}} \right) + \log \left( \frac{- 6 - 2\sqrt{34}}{- 6 + 2\sqrt{34}} \right) \right]\]
\[ = \frac{1}{\sqrt{34}} \log \left( \frac{6 + 2\sqrt{34}}{6 - 2\sqrt{34}} \times \frac{4 + 2\sqrt{34}}{4 - 2\sqrt{34}} \right)\]
\[ = \frac{1}{\sqrt{34}} \log \left( \frac{160 + 20\sqrt{34}}{160 - 20\sqrt{34}} \right)\]
\[ = \frac{1}{\sqrt{34}} \log \left( \frac{8 + \sqrt{34}}{8 - \sqrt{34}} \right)\]