Question

lf $\int f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\log \left(f\right(x\left)\right)+c$, then $f\left(x\right)$ is equal to

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

Answer 1

Answer: (A)

$\frac{2\sec 2x}{(b^2-a^2)}$

Given, $\int f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\log \left(f\right(x\left)\right)+c$,
Differentiating both sides w.r.t. $x$,
$f\left(x\right)\sin x\cos x=\frac{1}{2(b^2-a^2)}\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$
$=2(b^2-a^2)\sin x\cos x=\frac{f^{\prime }\left(x\right)}{{f\left(x\right)}^2}$
Integrating both sides w.r.t $x$, we get
$\Rightarrow (b^2-a^2)\int \sin 2xdx=\int \frac{f^{\prime }\left(x\right)}{{f\left(x\right)}^2}dx$
Substitute, $f\left(x\right)=t\Rightarrow f^{\prime }\left(x\right)dx=dt$
$\Rightarrow (b^2-a^2)\frac{-\cos 2x}{2}=\int t^{-2}dt$
$\Rightarrow (b^2-a^2)\frac{-\cos 2x}{2}=\frac{t^{-1}}{-1}$
$\Rightarrow (b^2-a^2)\frac{\cos 2x}{2}=\frac{1}{f\left(x\right)}$
$\Rightarrow f\left(x\right)=\frac{2}{(b^2-a^2)}\sec 2x$