Question

If $\int f(x)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\log f(x)+c$, 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, $\int f(x)\sin x\cos xdx=\frac{1}{2\left(b^2-a^2\right)}\log (f(x))+c$
$\Rightarrow f(x)\sin x\cos x=\frac{1}{2\left(b^2-a^2\right)}\cdot \frac{1}{f(x)}\cdot f^{\prime }(x)$
$\Rightarrow f(x)\sin 2x=\frac{1}{b^2-a^2}\cdot \frac{f^{\prime }(x)}{f(x)}$
$\Rightarrow \sin 2x=\frac{1}{b^2-a^2}\frac{f^{\prime }(x)}{(f(x){)}^2}$
$\Rightarrow \int \sin 2xdx=\frac{1}{b^2-a^2}\int \frac{f^{\prime }(x)}{(f(x){)}^2}dx$
$\Rightarrow \frac{-\cos 2x}{2}=\frac{1}{b^2-a^2}\cdot \left(\frac{-1}{f(x)}\right)$
$\Rightarrow \frac{\cos 2x\left(b^2-a^2\right)}{2}=\frac{1}{f(x)}$
$\Rightarrow f(x)=\frac{2}{\left(b^2-a^2\right)\cos 2x}$