Question

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

Answer 1

$f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}logf\left(x\right)+c$

Differentiating both sides

$f\left(x\right)\sin x\cos x=\frac{f^{\prime }\left(x\right)}{2(b^2-a^2)f\left(x\right)}$

$\frac{f^{\prime }\left(x\right)}{\left(f\right(x){)}^2}=(b^2-a^2)\sin 2x$

$\int f^{\prime }\left(x\right)\left(f\right(x){)}_{dx}^{-2}=\int (b^2-a^2)\sin 2xdx$

$-\frac{1}{f\left(x\right)}=-\frac{c}{2}\cos 2x+c^{\prime }$.