Question

Evaluate: (i) $\int \sqrt{\frac{1+\cos 2x}{2}}dx$
(ii) $\int \sqrt{\frac{1-\cos 2x}{2}}dx$

Answer 1

(i)
$$\begin{gathered} \int \sqrt{\frac{1+\cos 2x}{2}}dx \\ \int \sqrt{\frac{2{\cos }^{2}x}{2}}dx\left[\therefore 1+\cos 2A=2{\cos }^{2}A\right] \\ =\int \cos xdx \\ =\sin x+C \end{gathered}$$

(ii)
$$\begin{gathered} \int \sqrt{\frac{1-\cos 2x}{2}}dx \\ =\int \sqrt{\frac{2{\sin }^{2}x}{2}}dx\left[\therefore 1-\cos 2x=2{\sin }^{2}x\right] \\ =\int \sin xdx \\ =-\cos x+C \end{gathered}$$