$A:\int \frac{1}{3+4\cos x}dx=\frac{1}{\sqrt{7}}\log \left|\frac{{\displaystyle \sqrt{7}+\tan (x/2)}}{{\displaystyle \sqrt{7}-\tan (x/2)}}\right|+c$

 $R:If a<b then$$\int \frac{dx}{a+b\cos x}=$$\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{{\displaystyle \sqrt{b+a}+\sqrt{b-a}\tan (x/2)}}{{\displaystyle \sqrt{b+a}-\sqrt{b-a}\tan (x/2)}}\right|+c$

$$\begin{gathered} R:a<b then \\ \int \frac{dx}{a+b\cos x} \\ =\int \frac{1}{\left[a+b\left({\displaystyle \frac{1-t^2}{1+t^2}}\right)\right]}\times \frac{2dt}{\left(1+t^2\right)} Put \tan \frac{x}{2}=t \\ =\int \frac{2dt}{a+a{t}^2+b-b{t}^2} se{c}^2\frac{x}{2} \frac{1}{2}dx=dt \\ dx=\frac{2dt}{1+t^2} \\ and \cos x=\frac{1-t^2}{1+t^2} \\ =\int \frac{2dt}{\left(a+b\right)-\left(b-a\right)t^2} \\ =2\int \frac{dt}{{\left(\sqrt{a+b}\right)}^2-{\left[\sqrt{\left(b-a\right)} t\right]}^2} \\ =\frac{2}{2\left(\sqrt{a+b}\right)}\frac{1}{\sqrt{b-a}}\log \left|\frac{\sqrt{a+b}+\sqrt{b-a} t}{\sqrt{a+b}-\sqrt{b-a} t}\right|+C \\ =\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{\sqrt{a+b}+\sqrt{b-a} \tan {\displaystyle \frac{x}{2}}}{\sqrt{a+b}-\sqrt{b-a} \tan {\displaystyle \frac{x}{2}}}\right|+C \\ Now a=3, b=4 \\ \Rightarrow \int \frac{1}{3+4\cos x}=\frac{1}{\sqrt{16-9}}\log \left|\frac{\sqrt{7}+\sqrt{1}\tan {\displaystyle \frac{x}{2}}}{\sqrt{7}-\sqrt{1}\tan \frac{x}{2}}\right| \\ =\frac{1}{\sqrt{7}}\log \left|\frac{\sqrt{7}+\sqrt{1}\tan {\displaystyle \frac{x}{2}}}{\sqrt{7}-\sqrt{1}\tan \frac{x}{2}}\right|+C \\ Both A, R are true and R is the correct explanation of A \end{gathered}$$