KCET · Maths · Functions
\( \int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x \) is equal to
- A \( 2(\sin x+x \cos \theta)+C \)
- B \( 2(\sin x-x \cos \theta)+C \)
- C \( 2(\sin x+2 x \cos \theta)+C \)
- D \( 2(\sin x-2 x \cos \theta)+C \)
Answer & Solution
Correct Answer
(A) \( 2(\sin x+x \cos \theta)+C \)
Step-by-step Solution
Detailed explanation
Given that, \(\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d\)
Since, \(\cos 2 \theta=2 \cos ^{2} \theta 1\) So,
\(\int \frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \theta-1\right)}{\cos x-\cos \theta} d x\)
\(=2 \int(\cos x+\cos \theta) d x=2(\sin x+x \cos \theta)+C\)
Since, \(\cos 2 \theta=2 \cos ^{2} \theta 1\) So,
\(\int \frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \theta-1\right)}{\cos x-\cos \theta} d x\)
\(=2 \int(\cos x+\cos \theta) d x=2(\sin x+x \cos \theta)+C\)
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