Inetgral

\hspace{-16}\bf{(a)::\int \sqrt{\sec x+1}dx = \int\frac{\sqrt{\sec x+1}\cdot \sqrt{\sec x-1}}{\sqrt{\sec x-1}}dx}$\\\\\\ $\bf{=\int\frac{\tan x}{\sqrt{\sec x-1}}dx = \int\frac{\sec x\cdot \tan x}{\sec x \cdot \sqrt{\sec x-1}}dx}$\\\\\\ Now $\bf{\sec x-1 = t^2\;,}$ Then $\bf{\sec x\cdot \tan x dx = 2tdt}$\\\\\\ So Integral is $\bf{\int\frac{2tdt}{\left(1+t^2\right)\cdot t}dt = 2\int\frac{1}{1+t^2}dt = 2\tan^{-1}(t)+\mathbb{C}}$\\\\\\ So $\bf{\int \sqrt{\sec x -1}dx = 2\tan^{-1}\left(\sqrt{\sec x-1}\right)+\mathbb{C}}$\\\\\\

\hspace{-16}\bf{(b)::\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx = \int\frac{2\cos \left(\frac{9x}{2}\right)\cdot \cos \left(\frac{x}{2}\right)}{1-2\left(2\cos^2 \frac{3x}{2}-1\right)}dx}$\\\\\\ $\bf{=\int\frac{2\cos \left(\frac{9x}{2}\right)\cdot \cos \left(\frac{x}{2}\right)}{3-4\cos^2\frac{3x}{2}}dx= -\int\frac{2\cos \left(\frac{3x}{2}\right)\cdot \cos \left(\frac{x}{2}\right)\cdot \cos \left(\frac{9x}{2}\right)}{4\cos^3\frac{3x}{2}-3\cos \left(\frac{3x}{2}\right)}dx}$\\\\\\ $\bf{=-\int\frac{\left\{\cos \left(2x\right)+\cos (x)\right\}\cdot \cos \left(\frac{9x}{2}\right) }{\cos \left(\frac{9x}{2}\right)}dx = -\frac{1}{2}\sin (2x)-\sin (x)+\mathbb{C}}$

\hspace{-20}\bf{(c)::\int \frac{f(x)\cdot \phi^{'}(x)+\phi(x)\cdot f^{'}(x)}{\left(f(x)\cdot \phi(x)+1\right)\cdot \sqrt{f(x)\cdot \phi(x)+1}}dx}$\\\\\\ Now Let $\bf{f(x)\cdot \phi(x)+1 = t^2,}$ Then $\bf{\left(f(x)\cdot \phi^{'}(x)+\phi(x)\cdot f^{'}(x)\right)dx = 2tdt}$\\\\\\ So Integral Convert into $\bf{=\int\frac{2t}{t^2\cdot t}dt = 2\int t^{-2}dt = -\frac{2}{t}+\mathbb{C}}$\\\\\\ So $\bf{\int \frac{f(x)\cdot \phi^{'}(x)+\phi(x)\cdot f^{'}(x)}{\left(f(x)\cdot \phi(x)+1\right)\cdot \sqrt{f(x)\cdot \phi(x)+1}}dx=-\frac{2}{\sqrt{f(x)\cdot \phi(x)+1}}+\mathbb{C}}$

answer will be 2cos(inverse)of(√cosx)+c

How???

thank you