Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{dx}{a\cos x+b\sin x}a,b>0$

Answer 1

$$\begin{gathered} {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{a\cos x+b\sin x}dx \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{a\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2\frac{x}{2}}}\right)+b\left({\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2\frac{x}{2}}}\right)}dx \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\left(1+{\tan }^2\frac{x}{2}\right)}{a-a{\tan }^2\frac{x}{2}+2b\tan {\displaystyle \frac{x}{2}}}dx \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{se{c}^2{\displaystyle \frac{x}{2}}}{a-ata{n}^2\frac{x}{2}+2btan{\displaystyle \frac{x}{2}}}dx \\ Let\tan \frac{x}{2}=t,Then,\frac{1}{2}se{c}^2\frac{x}{2}dx=dt \\ Whenx=0,t=0,x=\frac{\mathrm{\pi }}{2},t=1 \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ I={\int }_0^1\frac{2\mathrm{dt}}{\mathrm{a}-{\mathrm{at}}^2+2\mathrm{bt}} \\ ={\int }_0^1\frac{2dt}{-a\left[t^2-{\displaystyle \frac{2bt}{a}}-1\right]} \\ =\frac{2}{a}{\int }_0^1\frac{\mathrm{dt}}{-\left[{\left(t-{\displaystyle \frac{b}{a}}\right)}^2-1-{\displaystyle \frac{b^2}{a^2}}\right]} \\ =\frac{2}{a}{\int }_0^1\frac{dt}{\left({\displaystyle \frac{b^2}{a^2}}+1\right)-{\left(t-{\displaystyle \frac{b}{a}}\right)}^2} \\ =\frac{2}{a}\left[\frac{1}{2\sqrt{{\displaystyle \frac{a^2+b^2}{a^2}}}}{\left(\log \left|\frac{\sqrt{\frac{a^2+b^2}{a^2}}+\left(t-\frac{b}{a}\right)}{\sqrt{\frac{a^2+b^2}{a^2}}-\left(t-\frac{b}{a}\right)}\right|\right)}_0^1\right] \end{gathered}$$
$\frac{1}{\sqrt{a^2+b^2}}\log \left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)$