∫ sec x sin ( 2 x + θ ) + sin θ d x =

Given$I =\int \frac{\sec x}{\sqrt{\sin \left(2x+\theta \right)+\sin \theta }}dx$

Using the formula $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$\Rightarrow I =\int \frac{\sec x}{\sqrt{2\sin \left({\displaystyle \frac{2x+2\theta }{2}}\right)\cos \left({\displaystyle \frac{2x+\theta -\theta }{2}}\right)}}dx$

$\Rightarrow I =\int \frac{\sec x}{\sqrt{2\sin \left({\displaystyle x+\theta }\right)\cos x}}dx$

$\Rightarrow I =\int \frac{{\displaystyle \raisebox{1ex}{$\sec x$}\!\left/ \!\raisebox{-1ex}{$\cos x$}\right.}}{\left({\displaystyle \frac{\sqrt{{\displaystyle 2\sin \left(x+\theta \right)}}\sqrt{\cos x}}{\sqrt{\cos x}\sqrt{\cos x}}}\right)}dx$

$\Rightarrow I =\int \frac{1}{\sqrt{2}}\frac{{\displaystyle \raisebox{1ex}{$\sec x$}\!\left/ \!\raisebox{-1ex}{$\cos x$}\right.}}{\sqrt{{\displaystyle \frac{\sin x\cos \theta +\cos x\sin \theta }{\cos x}}}}dx$

use $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$ formula

$\Rightarrow I=\int \frac{1}{\sqrt{2}}\frac{{\sec }^{2}x}{\sqrt{\tan x\cos \theta +\sin \theta }}dx$

take$\tan x \cos \theta +\sin \theta =t$

$\Rightarrow {\sec }^{2}x \cos \theta dx=dt$

$\Rightarrow {\sec }^{2}x dx=\frac{dt}{\cos \theta }$

$\Rightarrow I=\frac{1}{\sqrt{2}}\int \frac{{\displaystyle \raisebox{1ex}{$dt$}\!\left/ \!\raisebox{-1ex}{$\cos \theta $}\right.}}{\sqrt{t}}$

$\Rightarrow I=\frac{1}{\sqrt{2}\cos \theta }\int t^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dt$

$\Rightarrow I=\frac{2}{\sqrt{2}\cos \theta }{t}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} +C$

$\Rightarrow I=\frac{\sqrt{2}}{\cos \theta }\sqrt{\tan x\cos \theta +\sin \theta }+C$

$\Rightarrow I=\sqrt{2}\sqrt{\frac{\tan x\cos \theta +\sin \theta }{{\cos }^2\theta }}+C$

$\Rightarrow I=\sqrt{\frac{2}{\cos \theta }}{\left(\tan x+\tan \theta \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C$

$\Rightarrow I=\sqrt{2 \sec \theta \left(\tan x+\tan \theta \right)}+C$