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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
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The correct answer is:
$f(x)=\sin x, A=\cos \alpha$
$\begin{aligned} & \text {} \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x=\int \frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha} d x \\ & =\int \cos x+\cos \alpha d x \\ & =\sin x+x \cos \alpha+c \\ & \therefore f(x)=\sin x \& A=\cos \alpha\end{aligned}$
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