Question

$\int \sqrt{secx-1}dx$ is equal to

• A. $2\ell n(cos\frac{x}{2}+\sqrt{{cos}^2\frac{x}{2}-\frac{1}{2})}+C$
• B. $\ell n(cos\frac{x}{2}+\sqrt{{cos}^2\frac{x}{2}-\frac{1}{2})}+C$
• C. $-2\ell n(cos\frac{x}{2}+\sqrt{{cos}^2\frac{x}{2}-\frac{1}{2})}+C$
• D. $none$ $of$ $these$

Answer 1

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

$\begin{array}{c}Accordingtoquestion........................\\ I=\sqrt{secx-1}dx\\ \Rightarrow \int \sqrt{\frac{1}{\cos x}-1}dx\\ \Rightarrow \int \sqrt{\frac{1-\cos x}{\cos x}}dx|\begin{array}{c}weknowthat:\\ \cos x=2{\cos }^2\frac{x}{2}-1\\ \cos x=1-2{\sin }^2\frac{x}{2}\\ \Rightarrow 1-\cos x=2{\sin }^2\frac{x}{2}\end{array}\\ I=\int \frac{\sqrt{2{\sin }^2\frac{x}{2}}}{\sqrt{2{\cos }^2\frac{x}{2}-1}}dx\\ =\int \frac{\sqrt{2}\sin \frac{x}{2}}{\sqrt{2{\cos }^2\frac{x}{2}-1}}dx|\begin{array}{c}let:\sqrt{2}\cos \frac{x}{2}=t\\ \Rightarrow \sqrt{2}-\sin \frac{x}{2}.\frac{1}{2}dx=dt,\\ \sin \frac{x}{2}dx=-\sqrt{2}dt\end{array}\\ I=\int \frac{\sqrt{2}.-\sqrt{2}dt}{\sqrt{t^2-1}}\\ \Rightarrow -2\int \frac{dt}{\sqrt{t^2-1}}\\ \Rightarrow -2\left[\log (\sqrt{t^2-1}+t\right]+c\\ \Rightarrow -2\left[\log \left(\sqrt{2{\cos }^2\frac{x}{2}-1}+\sqrt{2}\cos \frac{x}{2}\right)\right]+c\\ \Rightarrow -2\left[\log (\sqrt{2}\left(\sqrt{{\cos }^2\frac{x}{2}-\frac{1}{2}}+\cos \frac{x}{2}\right)\right]+c\\ \Rightarrow -2\left[\log (\sqrt{2}\left(\sqrt{{\cos }^2\frac{x}{2}-\frac{1}{2}}+\cos \frac{x}{2}\right)\right]+c\\ \Rightarrow -2\log \sqrt{2}-2\log \left[\cos \frac{x}{2}+\sqrt{{\cos }^2\frac{x}{2}-\frac{1}{2}}\right]+c|constant(-2\log \sqrt{2})=c_1\\ I=-2\log \left[\cos \frac{x}{2}+\sqrt{{\cos }^2\frac{x}{2}-\frac{1}{2}}\right]+c+c_1|constant(c+c_1)=C\\ I=-2\log \left[\cos \frac{x}{2}+\sqrt{{\cos }^2\frac{x}{2}-\frac{1}{2}}\right]+C\\ So,thatthecorrectoptionisC.\\ \end{array}$