Question

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

Answer 1

\[\int_0^\frac{\pi}{2} \frac{1}{4\cos x + 2\sin x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{4 - 4 \tan^2 \frac{x}{2} + 4\tan\frac{x}{2}} d x\]
\[\text{Let }\tan\frac{x}{2} = t,\text{ then }\frac{1}{2}se c^2 \frac{x}{2} dx = dt\]
\[\text{When }x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
\[ = \frac{- 1}{4} \int_0^1 \frac{dt}{\left( t - \frac{1}{2} \right)^2 - \frac{5}{4}}\]
\[ = \frac{- 1}{4} \times \frac{- 4}{\sqrt{5}} \left[ \log\frac{2t - 1 - \sqrt{5}}{2t - 1 + \sqrt{5}} \right]_0^1 \]
\[ = \frac{1}{\sqrt{5}}\log\frac{\sqrt{5} + 1}{\sqrt{5} - 1}\]