$\int \frac{\sec x}{\sqrt{\log (\sec x+ an x)}} d x=$

Question

$\int \frac{\sec x}{\sqrt{\log (\sec x+	an x)}} d x=$

Aicharya · Accepted Answer

Let $I=\int \frac{\sec x}{\sqrt{\log (\sec x+	an x)}} d x$ Put $\log (\sec x+	an x)=t \Rightarrow \frac{1}{\sec x+	an x}\left(\sec x 	an x+\sec ^{2} xight) d x=d t$ $\begin{aligned} & \frac{\sec x(\sec x+	an x)}{\sec x+	an x} \mathrm{dx}=\mathrm{dt} \Rightarrow \sec x \mathrm{dx}=\mathrm{dt} \ 	herefore \quad \mathrm{I} &=\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\int \mathrm{t}^{-\frac{1}{2}} \mathrm{dt}=\frac{\mathrm{t}^{\frac{1}{2}}}{\left(\frac{1}{2}ight)}+\mathrm{c} \ &=2 \sqrt{\mathrm{t}}+\mathrm{c}=2 \sqrt{\log (\sec x+	an x)}+\mathrm{c} \end{aligned}$

$\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x=$

Solution