Question

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

Answer 1

\[\int_0^\frac{\pi}{2} \frac{1}{a\cos x + b \sin x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{a\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + b\left( \frac{2\tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{a - a \tan^2 \frac{x}{2} + 2b \tan\frac{x}{2}}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{se c^2 \frac{x}{2}}{a - ata n^2 \frac{x}{2} + 2b tan\frac{x}{2}}dx\]
\[Let\ \tan\frac{x}{2} = t, Then, \frac{1}{2}se c^2 \frac{x}{2}dx = dt\]
\[When\ x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 \frac{2dt}{a - {at}^2 + 2bt}\]
\[ = \int_0^1 \frac{2dt}{- a\left[ t^2 - \frac{2bt}{a} - 1 \right]}\]
\[ = \frac{2}{a} \int_0^1 \frac{dt}{- \left[ \left( t - \frac{b}{a} \right)^2 - 1 - \frac{b^2}{a^2} \right]}\]
\[ = \frac{2}{a} \int_0^1 \frac{dt}{\left( \frac{b^2}{a^2} + 1 \right) - \left( t - \frac{b}{a} \right)^2}\]
\[ = \frac{2}{a}\left[ \frac{1}{2\sqrt{\frac{a^2 + b^2}{a^2}}} \left( \log\left| \frac{\sqrt{\frac{a^2 + b^2}{a^2}} + \left( t - \frac{b}{a} \right)}{\sqrt{\frac{a^2 + b^2}{a^2}} - \left( t - \frac{b}{a} \right)} \right| \right)_0^1 \right]\]