Question

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

• A. ${\tan }^{-1}(2\tan x+1)+c$
• B. ${\tan }^{-1}(\tan x+\frac{1}{2})+c$
• C. $\frac{1}{8}{\tan }^{-1}(\tan x+\frac{1}{2})+c$
• D. $\frac{1}{4}\tan (2\tan x+1)+c$

Answer 1

The correct option is C $\frac{1}{8}{\tan }^{-1}(\tan x+\frac{1}{2})+c$

$\int \frac{dx}{4si{n}^{2}x+4sinxcosx+5co{s}^{2}x}$

$\int \frac{dtanx}{(4ta{n}^{2}x+4tanx+1)+4}$

$=\int \frac{dtanx}{(2tanx+1{)}^2+4}$

$=\frac{1}{4\times 2}ta{n}^{-1}\left(\frac{2tanx+1}{2}\right)+c$

$=\frac{1}{8}ta{n}^{-1}(tanx+\frac{1}{2})+c$