Question

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

• 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

The correct option is C $\frac{2}{(b^2-a^2)\cos 2x}$

$\int f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\log f\left(x\right)+c$
differentiating with respect to $x$
$f\left(x\right)\sin x\cos x=\frac{1}{2(b^2-a^2)}\cdot \frac{1}{f\left(x\right)}\cdot f^{\prime }\left(x\right)f(x{)}^2=\frac{1}{(b^2-a^2)\sin 2x}{f}^{\prime }(x)(b^2-a^2)\sin 2x=\frac{1}{{f\left(x\right)}^2}{f}^{\prime }\left(x\right)$
integrating both the sides
$-(b^2-a^2)\frac{\cos 2x}{2}=-\frac{1}{f\left(x\right)}\Rightarrow f\left(x\right)=\frac{2}{(b^2-a^2)\cos 2x}+c$