Q103 / 180 · KCET 2022 · Mathematics
$\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x$ is equal to
- A$2(\sin x-x \cos \alpha)+c$
- Correct answerB$2(\sin x+x \cos \alpha)+c$
- C$2(\sin x-2 x \cos \alpha)+c$
- D$2(\sin x+2 x \cos \alpha)+c$
Solution
$$
\begin{aligned}
& L=1=\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x \\
& I=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x \\
& I=2 \int \frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha} d x
\end{aligned}
$$
$$
\begin{aligned}
& =2 \int \frac{(\cos x-\cos (x)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)} d x \\
& =2 \int(\cos x+\cos \alpha) d x=2(\sin x+x \cos \alpha)+c
\end{aligned}
$$
Hence, the correct answer is option: 'B'.