Question

$\int \sqrt{\sec x-1}dx$ is equal to
(where $C$ is integration constant)

• A. $-\ln |(\cos x+\frac{1}{2})+\sqrt{{\cos }^{2}x+\cos x}|+C$
• B. $-\ln |(\sin x+\frac{1}{2})+\sqrt{2{\sin }^{2}x+\sin x}|+C$
• C. $-\ln |(\cos x+\frac{1}{2})+\sqrt{2{\sin }^{2}x+\cos x}|+C$
• D. $-\ln |(\sin x+\frac{1}{2})+\sqrt{{\cos }^{2}x+2\cos x}|+C$

Answer 1

The correct option is A $-\ln |(\cos x+\frac{1}{2})+\sqrt{{\cos }^{2}x+\cos x}|+C$

$I=\int \sqrt{\sec x-1}dx=\int \sqrt{\frac{1-\cos x}{\cos x}}dx=\int \sqrt{\frac{(1-\cos x)}{\cos x}\times \frac{(1+\cos x)}{(1+\cos x)}}dx=\int \sqrt{\frac{1-{\cos }^{2}x}{\cos x+{\cos }^{2}x}}dx=\int \frac{\sin x}{\sqrt{{\cos }^{2}x+\cos x}}dx$
Putting $\cos x=t$
$\Rightarrow -\sin xdx=dt\therefore I=\int \frac{-dt}{\sqrt{t^2+t}}=-\int \frac{dt}{\sqrt{{(t+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}}=-\ln |(t+\frac{1}{2})+\sqrt{{(t+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}|+C=-\ln |(t+\frac{1}{2})+\sqrt{t^2+t}|+C\Rightarrow I=-\ln |(\cos x+\frac{1}{2})+\sqrt{{\cos }^{2}x+\cos x}|+C$