Question

$\int \frac{\cos x}{{\displaystyle \frac{1}{4}}-{\cos }^{2}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\cos x}{{\displaystyle \frac{1}{4}}-{\cos }^{2}x}dx \\ =\int \frac{\cos x}{{\displaystyle \frac{1}{4}}-\left(1-{\sin }^{2}x\right)}dx \\ =\int \frac{\cos x}{{\sin }^{2}x-{\displaystyle \frac{3}{4}}}dx \\ =\int \frac{\cos x}{{\sin }^{2}x-{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dx \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \therefore I=\int \frac{1}{t^2-{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dt \\ =\frac{1}{2\times {\displaystyle \frac{\sqrt{3}}{2}}}\ln \left|\frac{t-{\displaystyle \frac{\sqrt{3}}{2}}}{t+{\displaystyle \frac{\sqrt{3}}{2}}}\right|+C \\ =\frac{1}{\sqrt{3}}\ln \left|\frac{2t-\sqrt{3}}{2t+\sqrt{3}}\right|+C \\ =\frac{1}{\sqrt{3}}\ln \left|\frac{2\sin x-\sqrt{3}}{2\sin x+\sqrt{3}}\right|+C\mathit{}\left[\because \mathit{}t=\sin x\right] \end{gathered}$$