Integral Calculus Question 392
Question: $ \int_{{}}^{{}}{\frac{\sqrt{\tan x}}{\sin x\cos x}}\ dx= $
[Bihar CEE 1974; MP PET 2002; Kerala (Engg.) 2002]
Options:
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 $
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Answer:
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 $ .
BETA
