Question

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

• A. $\frac{2}{(b^2 - a^2) \sin \, 2x} $
• B. $\frac{2}{ab\, \sin \, 2x}$
• C. $\frac{2}{(b^2 - a^2) \cos \, 2x} $
• D. $\frac{2}{ab \, \cos \, 2x}$

Answer 1

Answer: (C)

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