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
- A \(\frac{2}{\left(b^{2}-a^{2}\right) \sin 2 x}\)
- B \(\frac{2}{a b \sin 2 x}\)
- C \(\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}\)
- D \(\frac{2}{a b \cos 2 x}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}\)
Step-by-step Solution
Detailed explanation
We have. \(\int f(x) \sin x \cos x d x=\frac{1}{2 b^{2}-a^{2}} \log (f(x))+c\) \(\Rightarrow f(x) \sin x \cos x=\frac{1}{\left.2 b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)\) \(\Rightarrow \quad f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}\)…
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