Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1}{a+b\cos x}dx=$

(a) $\frac{\mathrm{\pi }}{\sqrt{a^2-b^2}}$

(b) $\frac{\mathrm{\pi }}{ab}$

(c) $\frac{\mathrm{\pi }}{a^2+b^2}$

(d) (a + b) π

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\frac{\mathrm{\pi }}{\sqrt{a^2-b^2}} \\ \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\mathrm{\pi }}\frac{1}{a+b\cos x}\mathit{d}x \\ ={\int }_0^{\mathrm{\pi }}\frac{1}{a+b{\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}\mathit{d}x \end{gathered}$$

$$\begin{gathered} ={\int }_0^{\mathrm{\pi }}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{a\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)+b\left(1-{\tan }^2\frac{x}{2}\right)}\mathit{d}x \\ ={\int }_0^{\mathrm{\pi }}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{\left(a+b\right)+\left(a-b\right){\tan }^2{\displaystyle \frac{x}{2}}}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{\left(a+b\right)+\left(a-b\right){\tan }^2{\displaystyle \frac{x}{2}}}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \mathit{\Rightarrow }{\sec }^2\frac{x}{2}dx=2dt \\ \mathrm{When}x\to 0;t\to 0 \\ \mathrm{and}x\to \mathrm{\pi };t\to \infty \\ \therefore I={\int }_0^{\infty }\frac{2dt}{\left(a+b\right)+\left(a-b\right)t^2} \\ =\frac{2}{a-b}{\int }_0^{\infty }\frac{1}{{\displaystyle \left(\frac{a+b}{a-b}\right)+t^2}}dt \end{gathered}$$

$$\begin{gathered} =\frac{2}{\left(a-b\right)}{\int }_0^{\infty }\frac{1}{{\displaystyle {\left(\sqrt{\frac{a+b}{a-b}}\right)}^2+t^2}}dt \\ =\frac{2}{\left(a-b\right)}\times \sqrt{\frac{a-b}{a+b}}{\left[{\tan }^{-1}\frac{t}{\sqrt{{\displaystyle \frac{a+b}{a-b}}}}\right]}_0^{\infty } \\ =\frac{2}{\sqrt{a^2-b^2}}\left[\frac{\mathrm{\pi }}{2}-0\right] \\ =\frac{2}{\sqrt{a^2-b^2}}\left[\frac{\mathrm{\pi }}{2}\right] \\ =\frac{\mathrm{\pi }}{\sqrt{a^2-b^2}} \end{gathered}$$