Question

$\int \frac{dx}{4{\sin }^{2}x+4\sin x\cos x+5{\cos }^{2}x}$ is equal to

• A. ${\tan }^{-1}\left(\tan x+\frac{1}{2}\right)+c$
• B. $\frac{1}{4}{\tan }^{-1}\left(\tan x+\frac{1}{2}\right)+c$
• C. $4{\tan }^{-1}\left(\tan x+\frac{1}{2}\right)+c$
• D. none of these

Answer 1

The correct option is B $\frac{1}{4}{\tan }^{-1}\left(\tan x+\frac{1}{2}\right)+c$

$\int \frac{dx}{4{\sin }^{2}x+4\sin x\cos x+5{\cos }^{2}x}$

$=\int \frac{\frac{1}{{\cos }^{2}x}dx}{\frac{4{\sin }^{2}x}{{\cos }^{2}x}+\frac{4\sin x\cos x}{{\cos }^{2}x}+\frac{5{\cos }^{2}x}{{\cos }^{2}x}}$

$=\int \frac{{\sec }^{2}xdx}{4{\tan }^{2}x+4\tan x+5}$

$\tan x=R$

${\sec }^{2}xdx=dt$

$=\int \frac{dt}{4t^2+4t+5}$

$=\frac{1}{4}\int \frac{dt}{(t+1{)}^2+(\frac{1}{2}{)}^2}$

$=\frac{1}{4}{\tan }^{-1}\left(\frac{t+\frac{1}{2}}{1}\right)$

$=\frac{1}{4}{\tan }^{-1}\left(\frac{2\tan +1}{2}\right)+c$

or, $=\frac{1}{4}{\tan }^{-1}(\tan x+\frac{1}{2})+c$