Q. ${\int_{}^{}\left\{ 1 + 2{tanx(tan}{x + \sec{x)}} \right\}}^{\frac{1}{2}}dx$ is equal to
A) $\log{(\sec{x + \tan{x) + c}}}$
B) ${\log{(\sec{x + \tan{x)}}}}^{1/2} + c$
C) $\log{\sec{x(\sec{x + \tan{x) + c}}}}$
D) None of the above
Explanation
(c)
Let $I = {\int_{}^{}\left\{ 1 + 2{tanx(tan}{x + \sec{x)}} \right\}}^{\frac{1}{2}}dx$
$= \int_{}^{}\left\{ 1 + 2{\tan^{2}x + 2\tan}{x\sec x} \right\}^{1/2}dx$
$= \int_{}^{}\left\{ \sec^{2}x + \tan^{2}{x + 2\tan{x\sec x}} \right\}^{1/2}dx$
$= \int_{}^{}{(\sec{x + \tan{x)}}}dx$
$= \log{(\sec{x + \tan{x)}}} + \log{\sec x} + c$
$= \log{\sec{x(\sec{x + \tan{x) + c}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.