Question

Let $I=\int \frac{dx}{a+b\cos x}$ where $a,b>0$ and $a+b={\lambda }_1,a-b={\lambda }_2$ (where $C$ is a constant of integration)

• A. If ${\lambda }_2=0$, then $I=\frac{2}{{\lambda }_1}\tan \left(\frac{x}{2}\right)+C$
• B. If ${\lambda }_2>0$, then $I=\frac{1}{\sqrt{{\lambda }_1{\lambda }_2}}{\tan }^{-1}(\sqrt{\frac{{\lambda }_2}{{\lambda }_1}}\tan \frac{x}{2})+C$
• C. If ${\lambda }_2<0$, then $I=\frac{2}{\sqrt{-{\lambda }_1{\lambda }_2}}{\log }_e|\frac{\sqrt{{\lambda }_1}+\sqrt{-{\lambda }_2}\tan \frac{x}{2}}{\sqrt{{\lambda }_1}-\sqrt{-{\lambda }_2}\tan \frac{x}{2}}|+C$
• D. If ${\lambda }_2<0$, then $I=\frac{1}{\sqrt{-{\lambda }_1{\lambda }_2}}lo{g}_e|\frac{\sqrt{{\lambda }_1}+\sqrt{-{\lambda }_2}\tan \frac{x}{2}}{\sqrt{{\lambda }_1}-\sqrt{-{\lambda }_2}\tan \frac{x}{2}}|+C$

Answer 1

Answer: (A), (D)

If ${\lambda }_2<0$, then $I=\frac{1}{\sqrt{-{\lambda }_1{\lambda }_2}}lo{g}_e|\frac{\sqrt{{\lambda }_1}+\sqrt{-{\lambda }_2}\tan \frac{x}{2}}{\sqrt{{\lambda }_1}-\sqrt{-{\lambda }_2}\tan \frac{x}{2}}|+C$

$I=\int \frac{dx}{a+b(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}})}\tan \left(\frac{x}{2}\right)=t\Rightarrow \frac{dt}{dx}=\frac{1}{2}{\sec }^2\frac{x}{2}$
$I=\int \frac{2dt}{(a+b)+(a-b)t^2}=2\int \frac{dt}{{\lambda }_1+{\lambda }_2{t}^2}\text{given that}{\lambda }_1>0\text{Case (i) If}{\lambda }_2>0I=\frac{2}{\sqrt{{\lambda }_1{\lambda }_2}}{\tan }^{-1}(\sqrt{\frac{{\lambda }_2}{\lambda }}.t)+C$
$\text{Case (ii)}{\lambda }_2=0\Rightarrow I=2\int \frac{dt}{{\lambda }_1}=\frac{2}{{\lambda }_1}\tan \frac{x}{2}+C$
$\text{Case (iii)}{\lambda }_2<0\Rightarrow I=2\int \frac{dt}{{\lambda }_1-(-{\lambda }_2)t^2}\Rightarrow I=\frac{1}{\sqrt{-{\lambda }_1{\lambda }_2}}\ln |\frac{\sqrt{{\lambda }_1}+\sqrt{-{\lambda }_2}\tan \frac{x}{2}}{\sqrt{{\lambda }_1}-\sqrt{-{\lambda }_2}\tan \frac{x}{2}}|+C$