If int frac{sin ^2 alpha-sin ^2 x}{cos x-cos | vedclass

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(C) We are given the integral $I = \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x$. Using the identity $\sin^2 	heta = 1 - \cos^2 	het

Vedclass · Accepted Answer

$f(x)=\sin x, A=\cos \alpha$

Vedclass · Answer

(C) We are given the integral $I = \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x$. Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we can rewrite the numerator: $\sin^2 \alpha - \sin^2 x = (1 - \cos^2 \alpha) - (1 - \cos^2 x) = \cos^2 x - \cos^2 \alpha$. Substituting this into the integral,we get: $I = \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} d x$. Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$: $I = \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} d x$. Canceling the common term $(\cos x - \cos \alpha)$: $I = \int (\cos x + \cos \alpha) d x$. Integrating with respect to $x$: $I = \sin x + x \cos \alpha + C$. Comparing this with $f(x) + Ax + B$,we identify $f(x) = \sin x$ and $A = \cos \alpha$.

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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