Explanation

(b)
$\int_{}^{}\frac{\sec{x\ dx}}{\sqrt{\sin{(2x + \theta) + \sin\theta}}}$
$= \int_{}^{}\frac{\sec{x\ dx}}{\sqrt{2\sin{(x + \theta)\cos x}}}$
$= \frac{1}{\sqrt{2}}\int_{}^{}\frac{{\sec^{3/2}x}{dx}}{\sqrt{\sin{x\cos{\theta + \sin{\theta\cos x}}}}}$
$= \frac{1}{\sqrt{2}}\int_{}^{}\frac{\sec^{2}{x\ dx}}{\sqrt{\tan{x\cos{\theta + \sin\theta}}}}$
Put $\tan{x = t\ \Rightarrow \sec^{2}{x\ dx} = dt\ }$
$\therefore I = \frac{1}{\sqrt{2}}\int_{}^{}\frac{dt}{\sqrt{t\cos{\theta + \sin\theta}\ }}$
$= \frac{2}{\sqrt{2}} \bullet \frac{\sqrt{t\cos{\theta + \sin\theta}}}{\cos\theta} + c$
$= \sqrt{2(\tan{x\sec{\theta + \tan{\theta\sec{\theta)}}}}} + c$