A) \[{{\sin }^{-1}}(\tan x)\]
B) \[\tan x\]
C) \[{{\cos }^{-1}}(\tan x)\]
D) \[\frac{\sin x}{\sqrt{\cos x}}\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{\sec x\,dx}{\sqrt{\cos 2x}}}=\int_{{}}^{{}}{\frac{\sec x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}}\,dx\] \[=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{\sqrt{1-{{\tan }^{2}}x}}}\] {Multiplying \[N'r\] and \[D'r\] by \[\sec x\}\] Now putting \[\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt,\] we get the integral \[={{\sin }^{-1}}t={{\sin }^{-1}}(\tan x).\] Trick : Since \[\frac{d}{dx}\{{{\sin }^{-1}}(\tan x)\}=\frac{{{\sec }^{2}}x}{\sqrt{1-{{\tan }^{2}}x}}\] \[=\frac{{{\sec }^{2}}x.\cos x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}=\frac{\sec x}{\sqrt{\cos 2x}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
