Question

$\int \frac{dx}{2+sin2x+cos2x}$ equals to

• A. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{1+tanx}{\sqrt{2}}\right)$
• B. $\frac{1}{2}ta{n}^{-1}\left(\frac{1+tanx}{2}\right)$
• C. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{1-tanx}{\sqrt{2}}\right)$
• D. $\sqrt{2}ta{n}^{-1}\left(\frac{\sqrt{2}}{1+tanx}\right)$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{1+tanx}{\sqrt{2}}\right)$

$\int \frac{dx}{2+sin2x+cos2x}$
Put tan x=t
$I=\int \frac{1}{2+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}.\frac{dt}{1+t^2}$
$=\int \frac{1}{t^2+2t+3}dt=\int \frac{1}{(t+1{)}^2+2}.dt$
$=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+c$
$=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{1+tanx}{\sqrt{2}}\right)+c$