Question

$\int_{}^{}\sqrt{\sec x–1}$ dx is equal to –

• A. 2 log $\left( \cos\frac{x}{2} + \sqrt{\cos^{2}\frac{x}{2}–\frac{1}{2}} \right) + C$
• B. log $\left( \cos\frac{x}{2} + \sqrt{\cos^{2}\frac{x}{2}–\frac{1}{2}} \right) + C$
• C. – 2 log $\left( \cos\frac{x}{2} + \sqrt{\cos^{2}\frac{x}{2}–\frac{1}{2}} \right) + C$
• D. None of these

Answer 1

Answer: (C)

– 2 log $\left( \cos\frac{x}{2} + \sqrt{\cos^{2}\frac{x}{2}–\frac{1}{2}} \right) + C$

Answer 2

$\int \sqrt { \frac { 1 } { \cos x } - 1 } d x$ =

= =

Put = t

= – = $\frac { \mathrm { dt } } { \mathrm { dx } }$ = $\sqrt { 2 } \sin \left( \frac { x } { 2 } \right) d x$ = –2dt

So Integral becomes

= $- 2 \int \frac { \mathrm { dt } } { \sqrt { \mathrm { t } ^ { 2 } - 1 } }$ = –2 log $\left| \sqrt { \mathrm { t } ^ { 2 } - 1 } + \mathrm { t } \right|$ + C