$\int[1+2 an x( an x+\sec x)]^{\frac{1}{2}} d x=$

Question

$\int[1+2 	an x(	an x+\sec x)]^{\frac{1}{2}} d x=$

Aicharya · Accepted Answer

Let $\left.I=\int[1+2 	an x(	an x+\sec x)]^{1 / 2} d x=\int 1+2 	an ^2 x+2 	an x \sec xight)^{1 / 2} d x$ $\int\left[\left(1+	an ^2 xight)+	an ^2 x+2 \sec x 	an xight]^{1 / 2} d x=\int\left(\sec ^2 x+	an ^2 x+2 \sec x 	an xight)^{1 / 2} d x$ $\begin{aligned} & =\int\left[(\sec x+	an x)^2ight]^{1 / 2} d x=\int(\sec x+	an x) d x=\int \sec x d x+\int 	an x d x \ & =\log |\sec x+	an x|-\log |\cos x|+c \quad=\log \frac{|\sec x+	an x|}{|\cos x|}+c \ & =\log [\sec x(\sec x+	an x)]+c \end{aligned}$

$\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=$

Solution