Question

$\int_{}^{}\frac{\sqrt{\tan x}}{\sin x.\cos x}$dx

• A. $2\sqrt{\sec x} + c$
• B. $2\sqrt{\tan x} + c$
• C. $\frac{2}{\sqrt{\tan x}} + c$
• D. $\frac{2}{\sqrt{\sec x}}$+ c

Answer 1

Answer: (B)

$2\sqrt{\tan x} + c$

Answer 2

$\int_{}^{}{\frac{\sqrt{\tan x}}{\sin x.\cos x}dx}$ $= \int_{}^{}{\frac{\tan x}{\sqrt{\tan x}.\sin x\cos x}dx}$

$= \frac{\sin x.\sec x}{\sqrt{\tan x}\sin x\cos x}dx$ $= \int_{}^{}{\frac{\sec^{2}x}{\sqrt{\tan x}}dx}$

Put $t = \tan x$ $\Rightarrow dt = \sec^{2}xdx,$ then it reduces to, $\int_{}^{}{\frac{1}{\sqrt{t}}dt = 2t^{1/2} + c} = 2\sqrt{\tan x} + C$