A) \[2\sqrt{\sec x}+c\]
B) \[2\sqrt{\tan x}+c\]
C) \[\frac{2}{\sqrt{\tan x}}+c\]
D) \[\frac{2}{\sqrt{\sec x}}+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx}=\int_{{}}^{{}}{\frac{\tan x}{\sqrt{\tan x}\sin x\cos x}dx}\] \[=\int_{{}}^{{}}{\frac{\sin x\sec x}{\sqrt{\tan x}\sin x\cos x}\,dx}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{\sqrt{\tan x}}\,dx}\] Put \[t=\tan x\Rightarrow dt={{\sec }^{2}}x\,dx,\] then it reduces to \[\int_{{}}^{{}}{\frac{1}{\sqrt{t}}\,dt}=2{{t}^{1/2}}+c=2\sqrt{\tan x}+c\].
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