KCET · Maths · Indefinite Integration
\(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x\) is equal to
- A \(2(\sin x-x \cos \alpha)+c\)
- B \(2(\sin x+x \cos \alpha)+c\)
- C \(2(\sin x-2 x \cos \alpha)+c\)
- D \(2(\sin x+2 x \cos \alpha)+c\)
Answer & Solution
Correct Answer
(B) \(2(\sin x+x \cos \alpha)+c\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
& L=1=\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x \\
& I=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x \\
& I=2 \int \frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha} d x
\end{aligned}
\]
\[
\begin{aligned}
& =2 \int \frac{(\cos x-\cos (x)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)} d x \\
& =2 \int(\cos x+\cos \alpha) d x=2(\sin x+x \cos \alpha)+c
\end{aligned}
\]
\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}
\]
\[
\begin{aligned}
& =2 \int \frac{(\cos x-\cos (x)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)} d x \\
& =2 \int(\cos x+\cos \alpha) d x=2(\sin x+x \cos \alpha)+c
\end{aligned}
\]
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