AP EAMCET · Maths · Indefinite Integration
If \(\int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x=f(x)+A x+B\) and \(B \in R\), then
- A \(f(x)=2 \sin x, A=\cos \alpha\)
- B \(f(x)=2 \sin x, A=2 \cos \alpha\)
- C \(f(x)=\sin x, A=\cos \alpha\)
- D \(f(x)=\sin x, A=2 \cos \alpha\)
Answer & Solution
Correct Answer
(C) \(f(x)=\sin x, A=\cos \alpha\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {} \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x=\int \frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha} d x \\ & =\int \cos x+\cos \alpha d x \\ & =\sin x+x \cos \alpha+c \\ & \therefore f(x)=\sin x \& A=\cos \alpha\end{aligned}\)
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