Question

Evaluate the integral
${\int }_0^{\pi }\frac{dx}{a+bcosx}$ where $a>b$

• A. $\pi \sqrt{a^2-b^2}$
• B. $\pi ab$
• C. $\frac{\pi }{\sqrt{a^2+b^2}}$
• D. $\frac{\pi }{\sqrt{a^2-b^2}}$

Answer 1

The correct option is D $\frac{\pi }{\sqrt{a^2-b^2}}$

$I={\int }_0^{\pi }\frac{dx}{a+bcosx};I={\int }_0^{\pi }\frac{dx}{a-bcosx}$

$2I={\int }_0^{\pi }\frac{2a}{a^2-b^{2}co{s}^{2}x}$

$2I=2a{\int }_0^{\pi }\frac{dxtanx}{a^2(1+ta{n}^{2}x)-b^2}$

$2I=2a{\int }_0^{\pi }\frac{dxtanx}{a^{2}ta{n}^{2}x+(a^2-b^2)}$

$2I=\frac{2a}{\sqrt{a^2-b^2}}\frac{1}{a}ta{n}^{-1}{\left(\frac{atanx}{\sqrt{a^2-b^2}}\right)}_0^{\pi }$

$I=\frac{\pi }{\sqrt{a^2-b^2}}$