MHT CET · Maths · Indefinite Integration
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)
- A \(\sqrt{\tan x}+C\)
- B \(\frac{1}{2} \sqrt{\tan x}+C\)
- C \(\frac{3}{2} \sqrt{\tan x}+C\)
- D \(2 \sqrt{\tan x}+C\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{\tan x}+C\)
Step-by-step Solution
Detailed explanation
\(f(x)=\sqrt{\tan x}, g(x)=\sin x \cdot \cos x\)
Now,
\(\int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\) \(\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x\)
\(=2 \sqrt{\tan x}+C\)
Now,
\(\int \frac{f(x)}{g(x)} d x=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x=\int \frac{\sqrt{\tan x \cdot \sec ^2 x}}{\tan x} d x=\) \(\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x\)
\(=2 \sqrt{\tan x}+C\)
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