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…
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)
- $\sqrt{\tan x}+C$
- $\frac{1}{2} \sqrt{\tan x}+C$
- $\frac{3}{2} \sqrt{\tan x}+C$
- $2 \sqrt{\tan x}+C$
Solution
$f(x)=\sqrt{\tan x}, g(x)=\sin x \cdot \cos x$
Now,
$\begin{aligned} & \int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x \\ & =2 \sqrt{\tan x}+C\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 2)
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