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If $\int f(x) \sin x \cos x d x=\frac{1}{2\left(b^{2}-a^{2}\right)} \log f(x)+c$
where $c$ is the constant of integration, then $f(x)$ is equal to
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where $c$ is the constant of integration, then $f(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
We have. $\int f(x) \sin x \cos x d x=\frac{1}{2 b^{2}-a^{2}} \log (f(x))+c$
$\Rightarrow f(x) \sin x \cos x=\frac{1}{\left.2 b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)$
$\Rightarrow \quad f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}$
$\Rightarrow \quad \sin 2 x=\frac{1}{b^{2}-a^{2}} \frac{f(x)}{(f(x))^{2}}$
$\Rightarrow \quad \int \sin 2 x d x=\frac{1}{b^{2}-a^{2}} \int \frac{f(x)}{(f(x))^{2}} d x$
$\Rightarrow \quad \frac{-\cos 2 x}{2}=\frac{1}{b^{2}-a^{2}} \cdot\left(\frac{-1}{f(x)}\right)$
$\Rightarrow \quad \quad \frac{\cos 2 x\left(b^{2}-a^{2}\right)}{2}=\frac{1}{f(x)}$
$\Rightarrow \quad f(x)=\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
$\Rightarrow f(x) \sin x \cos x=\frac{1}{\left.2 b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)$
$\Rightarrow \quad f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}$
$\Rightarrow \quad \sin 2 x=\frac{1}{b^{2}-a^{2}} \frac{f(x)}{(f(x))^{2}}$
$\Rightarrow \quad \int \sin 2 x d x=\frac{1}{b^{2}-a^{2}} \int \frac{f(x)}{(f(x))^{2}} d x$
$\Rightarrow \quad \frac{-\cos 2 x}{2}=\frac{1}{b^{2}-a^{2}} \cdot\left(\frac{-1}{f(x)}\right)$
$\Rightarrow \quad \quad \frac{\cos 2 x\left(b^{2}-a^{2}\right)}{2}=\frac{1}{f(x)}$
$\Rightarrow \quad f(x)=\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
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