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Q.
If $\int f(x) \sin x \cos x \,dx =\frac{1}{2\left(b^{2}-a^{2}\right)} \log [f(x)]+c$ then $f(x)$ is equal to
ManipalManipal 2008
Solution:
Given, $\int f(x) \sin x \cos x d x=\frac{1}{2\left(b^{2}-a^{2}\right)} \log [f(x)]+c$
On differentiating both sides, we get
$f(x) \sin x \cos x=\frac{d}{d x-}\left(\frac{\log [f(x)]}{-2\left(b^{2}-a^{2}\right)}+c\right)$
$\Rightarrow f(x) \sin x \cos x=\frac{1}{2\left(b^{2}-a^{2}\right)} \cdot \frac{1}{-f(x)} f'(x)$
$\Rightarrow 2\left(b^{2}-a^{2}\right) \sin x \cos x=\frac{f'(x)}{\{f(x)]^{2}}$
On integrating both sides, we get
$\int\left(2 b^{2} \sin x \cos x-2 a^{2} \sin x \cos x\right) d x$
$=\int \frac{f'(x)}{f f(x)]^{2}} d x$
$\Rightarrow -b^{2} \cos ^{2} x-a^{2} \sin ^{2} x=-\frac{1}{f(x)}$
$\therefore f(x)=\frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x}$