Explanation

(a)
Given, $\int_{}^{}{f(x)}\sin x\cos xdx = \frac{1}{2(b^{2} - a^{2})}\log{\{ f(x)\}} + c$
On differentiating both sides, we get
$f(x)\sin{x\cos{x = \frac{1}{2\left( b^{2} - a^{2} \right)}}}.\frac{1}{f(x)}f^{'}(x)$
$\Rightarrow \ \ 2\left( b^{2} - a^{2} \right)\sin x\cos x = \frac{f^{'}(x)}{\left\lbrack f(x) \right\rbrack^{2}}$
$\Rightarrow \ \ \int_{}^{}{\left( 2b^{2}\sin x\cos x - 2a^{2}\sin x\cos x \right)dx = \int_{}^{}\frac{f^{'}(x)}{\left\lbrack f(x) \right\rbrack^{2}}}dx$
$I_{1} - I_{2} = \int_{}^{}\frac{f^{'}(x)}{\left\lbrack f(x) \right\rbrack^{2}}\$ …(i)
Where $I_{1} = \int_{}^{}{2b^{2}\sin x\cos xdx}\$and$\ \ I_{2} = \int_{}^{}{2a^{2}}\sin x\cos xdx$
Now, $I_{1} = 2b^{2}\ \sin x\cos xdx$
Put $\cos x = t\ \ \Rightarrow \ \ - \sin xdx = dt$
$\therefore\ \ I_{1} = - \int_{}^{}{2b^{2}tdt} = - b^{2}t^{2} = - b^{2}\cos^{2}x$
and $I_{2} = \int_{}^{}{2a^{2}}\sin x\cos xdx$
Put $\sin x = t\ \ \Rightarrow \ \cos xdx = t$
$\therefore\ I_{2} = \int_{}^{}{2a^{2}}tdt = a^{2}t^{2} = a^{2}\sin^{2}x$
$\therefore$ From Eq. (i),
$- b^{2}\cos^{2}x - a^{2}\sin^{2}x = - \frac{1}{f(x)}$
$\Rightarrow \ \ f(x) = \frac{1}{a^{2}\sin^{2}{x + b^{2}\cos^{2}x}}$