Question

If $\int \frac{dx}{2\sin x-\cos x+5}=A{\tan }^{-1}\frac{f\left(x\right)}{\sqrt{5}}+C,$ for a fixed value of $A$ and function $f\left(x\right)$. Then
(where $C$ is integration constant)

• A. $A=\frac{1}{\sqrt{5}}$
• B. $A=\sqrt{5}$
• C. $f\left(x\right)=3\tan \left(\frac{x}{2}\right)+1$
• D. $f\left(x\right)=2\tan \left(\frac{x}{2}\right)+1$

Answer 1

Answer: (A), (C)

$f\left(x\right)=3\tan \left(\frac{x}{2}\right)+1$

$I=\int \frac{dx}{2\sin x-\cos x+5}\{\because \sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}},\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\}$
Put
$\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}dx=2dt\Rightarrow \sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$
Now,
$I=\int \frac{dt}{3t^2+2t+2}=\frac{1}{3}\int \frac{dt}{{(t+\frac{1}{3})}^2+{\left(\frac{\sqrt{5}}{3}\right)}^2}=\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{3t+1}{\sqrt{5}}\right)+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\therefore I=\frac{1}{\sqrt{5}}{\tan }^{-1}\left[\frac{3\tan \frac{x}{2}+1}{\sqrt{5}}\right]+C$