Q109 / 180 · KCET 2010 · Mathematics
If $\int f(x) \sin x \cdot \cos x d x$
$=\frac{1}{2\left(b^{2}-a^{2}\right)} \log f(x)+c,$
where $c$ is the constant of integration, then $f(x)$ is
- A$\frac{2}{a b \cos 2 x}$
- Correct answerB$\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
- C$\frac{2}{a b \sin 2 x}$
- D$\frac{2}{\left(b^{2}-a^{2}\right) \sin 2 x}$
Solution
If $\int f(x) \cdot \sin x \cdot \cos x d x$
$=\frac{1}{2\left(b^{2}-a^{2}\right)} \log f(x)+c$
$\begin{aligned}
&\text { LHS } \frac{1}{2} \int \mathrm{f}(\mathrm{x}) \cdot(2 \sin \mathrm{x} \cos \mathrm{x}) \mathrm{dx} \\
&\quad=\frac{1}{2} \int \mathrm{f}(\mathrm{x}) \cdot \sin 2 \mathrm{x} \cdot \mathrm{dx} \\
&{\left[\mathrm{Here}, \mathrm{put} \mathrm{f}(\mathrm{x})=\frac{2}{\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \times \frac{1}{\cos 2 \mathrm{x}}\right]} \\
&=\frac{1}{2} \int \frac{2}{\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \cdot \frac{\sin 2 \mathrm{x}}{\cos 2 \mathrm{x}} \mathrm{dx}
\end{aligned}$
$\begin{aligned}
&=\frac{1}{\left(b^{2}-a^{2}\right)} \int \tan 2 x d x \\
&=\frac{1}{\left(b^{2}-a^{2}\right)} \cdot \frac{\log \sec 2 x}{2}+c_{1} \\
&=\frac{1}{2\left(b^{2}-a^{2}\right)} \log \left(\frac{1}{\cos 2 x}\right)+c_{1}
\end{aligned}$
$\begin{aligned} & {\left[\text { Put }_{1}=\mathrm{c}+\frac{1}{2\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \log \frac{2}{\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)}\right] } \\=& \frac{1}{2\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \cdot \log \left(\frac{1}{\cos 2 \mathrm{x}}\right) \\ &+\frac{1}{2\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \log \left(\frac{2}{\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)}\right)+\mathrm{c} \\=& \frac{1}{2\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right)} \log \left[\frac{2}{\left(\mathrm{~b}^{2}-\mathrm{a}^{2}\right) \cos 2 \mathrm{x}}\right]+\mathrm{c} \end{aligned}$
Hence,
$f(x)=\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
Hence, the correct answer is option: 'B'.