Question

${\int }_0^{\pi /2}\frac{1}{a+bcosx}dx=$, where $a>|b|$

• A. $\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\sqrt{\frac{a+b}{a-b}}$
• B. $\frac{2}{\sqrt{a^2-b^2}}co{t}^{-l}\sqrt{\frac{a-b}{a+b}}$
• C. $\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\sqrt{\frac{a-b}{a+b}}$
• D. $\frac{\pi }{\sqrt{a^2-b^2}}$

Answer 1

The correct option is C $\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\sqrt{\frac{a-b}{a+b}}$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{1}{a+b\cos x}dx$
$\tan \left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dt$
Gives $\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
$I={\int }_0^1\frac{1}{a+b\left(\frac{1-t^2}{1+t^2}\right)}\frac{2dt}{1+t^2}=\frac{2}{a+b}{\int }_0^1\frac{1}{\frac{t^2(a-b)}{a+b}+1}dt$
Substitute $\frac{t\sqrt{a-b}}{\sqrt{a+b}}=u\Rightarrow \frac{dt\sqrt{a-b}}{\sqrt{a+b}}=du$

$I=\frac{2}{\sqrt{a-b}\sqrt{a+b}}{\int }_0^{\frac{\sqrt{a-b}}{\sqrt{a+b}}}\frac{1}{1+u^2}du$

$={\left[\frac{2{\tan }^{-1}u}{\sqrt{a-b}\sqrt{a+b}}\right]}_0^{\frac{\sqrt{a-b}}{\sqrt{a+b}}}=\frac{2{\tan }^{-1}\frac{\sqrt{a-b}}{\sqrt{a+b}}}{\sqrt{a-b}\sqrt{a+b}}-0=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\sqrt{\frac{a-b}{a+b}}$