$\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{(1+ an x)(2+ an x)} d x=$

Question

$\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{(1+	an x)(2+	an x)} d x=$

Aicharya · Accepted Answer

Put $1+	an x=\mathrm{t} \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}$ When $x=0, \mathrm{t}=1$ and when $x=\frac{\pi}{4}, \mathrm{t}=2$ $\begin{aligned} 	herefore \quad & \int_0^{\pi / 4} \frac{\sec ^2 x}{(1+	an x)(2+	an x)} \mathrm{d} x \ & =\int_1^2 \frac{\mathrm{dt}}{\mathrm{t}(1+\mathrm{t})} \ & =\int_1^2 \frac{\mathrm{dt}}{\mathrm{t}}-\int_1^2 \frac{\mathrm{dt}}{1+\mathrm{t}} \ & =\left[\log \mathrm{t}-\log _{(1+\mathrm{t}}(1)ight]_1^2 \ & =\log _{\mathrm{e}} 2-\log _{\mathrm{e}} 3+\log _{\mathrm{e}} 2 \ & =\log _{\mathrm{e}}\left(\frac{4}{3}ight) \end{aligned}$

$\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{(1+\tan x)(2+\tan x)} d x=$

Solution