Question

${\int }_0^{\pi /4}\frac{{\sec }^{2}x}{(1+\tan x)(2+\tan x)}dx$ equals:

• A. ${\log }_e\frac{2}{3}$
• B. ${\log }_{e}3$
• C. $\frac{1}{2}{\log }_e\frac{4}{3}$
• D. ${\log }_e\frac{4}{3}$

Answer 1

The correct option is D ${\log }_e\frac{4}{3}$

$I={\int }_0^{\pi /4}\frac{{\sec }^{2}xdx}{(1+\tan x)(2+\tan x)}$

Let $\tan x=t$

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

$\therefore I={\int }_0^1\frac{dt}{(1+t)(2+t)}={\int }_0^1\frac{(2+t)-(1+t)}{(2+t)(1+t)}$

$={\int }_0^1\frac{1}{1+t}-\frac{1}{2+t}$

$=\left[\ln \right(1+t)-\ln (2+t){]}_0^1$

$=(\ln 2-\ln 1)-(\ln 3-\ln 2)$

$=2\ln 2-\ln 3$

$=\ln \left(4/3\right)=\left(D\right)$