Question

Evaluate : $\int \frac{\sqrt{\cos 2x}}{\cos x}dx$

• A. $\sqrt{2}si{n}^{-1}\left(\sqrt{2}\sin x\right)-ta{n}^{-1}\left(\frac{\sin x}{\sqrt{\cos 2x}}\right)+C$
• B. $\sqrt{2}si{n}^{-1}\left(\sqrt{2}\cos x\right)+ta{n}^{-1}\left(\frac{\sin x}{\sqrt{\cos 2x}}\right)+C$
• C. $\sqrt{2}si{n}^{-1}\left(\sqrt{2}\sin x\right)-co{t}^{-1}\left(\frac{\sin x}{\sqrt{\sin 2x}}\right)+C$
• D. None

Answer 1

Answer: (A)

$\sqrt{2}si{n}^{-1}\left(\sqrt{2}\sin x\right)-ta{n}^{-1}\left(\frac{\sin x}{\sqrt{\cos 2x}}\right)+C$

Consider, $I=\int \frac{\sqrt{\cos 2x}}{\cos x}dx$

$I=\int \cos x\frac{\sqrt{\cos 2x}}{(\cos x{)}^2}dx$

$I=\int \cos x\frac{\sqrt{1-2(\sin x{)}^2}}{1-(\sin x{)}^2}dx$

substitute $\sqrt{2}\sin x=\sqrt{2}\sin u\cos xdx=\cos udu,\sin x=\frac{\sin u}{\sqrt{2}}$

$=\int \left(\frac{\cos udu}{\sqrt{2}}\right)\cdot \frac{\sqrt{1-2(\sin u{)}^2}}{1-{\left(\frac{\sin u}{\sqrt{2}}\right)}^2}$

$=\int \frac{\sqrt{2}(\cos u{)}^2}{2-(\sin u{)}^2}du$

$=\sqrt{2}\int \frac{(\cos u{)}^2}{1+(\cos u{)}^2}du$

substitute $z=\tan u,du=\frac{dz}{z^2+1},(\cos u{)}^2=\frac{1}{z^2+1}$

$=\sqrt{2}\int \frac{\frac{1}{z^2+1}}{1+\frac{1}{z^2+1}}\frac{dz}{z^2+1}$

$=u\sqrt{2}-{\tan }^{-1}\left(\frac{\tan u}{\sqrt{2}}\right)$

$=\sqrt{2}{\sin }^{-1}\left(\sqrt{2}\sin x\right)-{\tan }^{-1}\left(\frac{\sin x}{\sqrt{\cos 2x}}\right)+C$