Question

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

• A. $\frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x+k}$
• B. $\frac{1}{a^2{\sin }^{2}x-b^2{\cos }^{2}x+k}$
• C. $\frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x+k}$
• D. $\frac{1}{a^2{\cos }^{2}x-b^2{\sin }^{2}x+k}$

Answer 1

Answer: (A)

$\frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x+k}$

Given $\int f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\ln f\left(x\right)+c$
Differentiating both sides, we get
$f\left(x\right)\sin x\cos x=\frac{1}{2(b^2-a^2)}\frac{1}{f\left(x\right)}{f}^{\prime }\left(x\right)$
$\frac{f^{\prime }\left(x\right)}{\left(f\right(x){)}^2}=2(b^2-a^2)\sin x\cos x$

Using : $sin2x=2sinxcosx$
$\frac{f^{\prime }\left(x\right)}{\left(f\right(x){)}^2}=(b^2-a^2)\sin 2x$
Integrating both sides w.r.t. $x$, we get
$\int \frac{f^{\prime }\left(x\right)}{\left(f\right(x){)}^2}dx=(b^2-a^2)\int \sin 2xdx$

Take $f\left(x\right)=t\Rightarrow f^{\prime }\left(x\right)dx=dt$

$\therefore \int \frac{dt}{t^2}=-\frac{b^2-a^2}{2}\cos 2x$

$\therefore -\frac{1}{t}=-\frac{b^2-a^2}{2}({\cos }^{2}x-{\sin }^{2}x)$

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

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

$\therefore \frac{1}{f\left(x\right)}=b^2{\cos }^{2}x+a^2{\sin }^{2}x-\frac{b^2+a^2}{2}=b^2{\cos }^{2}x+a^2{\sin }^{2}x+k$, where $k=-\frac{a^2+b^2}{2}$
$\Rightarrow f\left(x\right)=\frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x+k}$