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WBJEE · Maths · Indefinite Integration

If \(\int f(x) \sin x \cos x d x=\frac{1}{2\left(b^{2}-a^{2}\right)} \log f(x)+c\)
where \(c\) is the constant of integration, then \(f(x)\) is equal to

  1. A \(\frac{2}{\left(b^{2}-a^{2}\right) \sin 2 x}\)
  2. B \(\frac{2}{a b \sin 2 x}\)
  3. C \(\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}\)
  4. D \(\frac{2}{a b \cos 2 x}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}\)

Step-by-step Solution

Detailed explanation

We have. \(\int f(x) \sin x \cos x d x=\frac{1}{2 b^{2}-a^{2}} \log (f(x))+c\) \(\Rightarrow f(x) \sin x \cos x=\frac{1}{\left.2 b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)\) \(\Rightarrow \quad f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}\)…