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If $f(x)=\sqrt{\tan x}$ and $g(x)=\sin x \cdot \cos x$, then $\int \frac{f(x)}{g(x)} d x$ is equal to (where $C$ is a constant of integration)
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Verified Answer
The correct answer is:
$2 \sqrt{\tan x}+C$
$f(x)=\sqrt{\tan x}, g(x)=\sin x \cdot \cos x$
Now,
$\begin{aligned} & \int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x \\ & =2 \sqrt{\tan x}+C\end{aligned}$
Now,
$\begin{aligned} & \int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x \\ & =2 \sqrt{\tan x}+C\end{aligned}$
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