Question

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

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

Answer 1

Answer: (A)

Given, $\int f(x)\sin x\cos xdx=\frac{1}{2\left(b^2-a^2\right)}\log [f(x)]+c$
On differentiating both sides, we get
$f(x)\sin x\cos x=\frac{d}{dx-}\left(\frac{\log [f(x)]}{-2\left(b^2-a^2\right)}+c\right)$
$\Rightarrow f(x)\sin x\cos x=\frac{1}{2\left(b^2-a^2\right)}\cdot \frac{1}{-f(x)}{f}^{\prime }(x)$
$\Rightarrow 2\left(b^2-a^2\right)\sin x\cos x=\frac{f^{\prime }(x)}{\{f(x){]}^2}$
On integrating both sides, we get
$\int \left(2b^2\sin x\cos x-2a^2\sin x\cos x\right)dx$
$=\int \frac{f^{\prime }(x)}{ff(x){]}^2}dx$
$\Rightarrow -b^2{\cos }^{2}x-a^2{\sin }^{2}x=-\frac{1}{f(x)}$
$\therefore f(x)=\frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}$