Mathematics

Solve $$\displaystyle\int\dfrac {\sqrt{\tan x}}{\sin x \cos x}dx$$


ANSWER

$$I=2\sqrt{\tan x}+C$$


SOLUTION

Consider the given integral.

$$ I=\int{\dfrac{\sqrt{\tan x}}{\sin x\cos x}dx} $$

$$ I=\int{\dfrac{\sqrt{\tan x}}{\tan x{{\cos }^{2}}x}dx} $$

$$ I=\int{\dfrac{{{\sec }^{2}}x}{\sqrt{\tan x}}dx} $$

 

Let $$t=\tan x$$

$$dt={{\sec }^{2}}xdx$$

 

Therefore,

$$ I=\int{\dfrac{dt}{\sqrt{t}}} $$

$$ I=2\sqrt{t}+C $$

 

On putting the value of $$t$$, we get

$$I=2\sqrt{\tan x}+C$$

 

Hence, this is the answer.

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