Question

If $\int \frac{\sin x}{{\sin }^{2}x+4{\cos }^{2}x}dx=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{g\left(x\right)}{\sqrt{3}}\right)+c$ then $g\left(x\right)=$

• A. $\sec x$
• B. $3\tan x$
• C. $\sin x$
• D. $3\cos x$

Answer 1

The correct option is D $3\cos x$

Let $I=\int \frac{\sin x}{{\sin }^{2}x+4{\cos }^{2}x}dx$

Let $I=\int \frac{\sin x}{{\cos }^{2}x-1+4{\cos }^{2}x}dx$

$\Rightarrow I=\int \frac{\sin xdx}{1+3{\cos }^{2}x}$

Putting $\cos x=t$ $\Rightarrow -\sin xdx=dt$

$\Rightarrow I=-\int \frac{dt}{1+3t^2}$

$\Rightarrow I=\frac{-1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}t}{1}\right)+c$

$\Rightarrow I=-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}\cos x}{1}\right)+c$

$\Rightarrow I=-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{3\cos x}{\sqrt{3}}\right)+c$

Comparing with the given equation. We get,

$g\left(x\right)=3\cos x$