Question

If $$\displaystyle \int\frac{\sin^{2}\alpha-\sin^{2}x}{\cos x-\cos\alpha}dx=f(x)+Ax+B$$, then

A
f(x)=2sinx, A=cosα
B
f(x)=2sinx, A=2cosα
C
f(x)=sinx, A=cosα
D
f(x)=sinx, A=2cosα

Solution

The correct option is C $$f(x)=\sin x,\ A=\cos\alpha$$$$\displaystyle \int \dfrac {\sin^2\alpha-\sin^2 x}{\cos x-\cos \alpha}\cdot dx$$$$\because \sin^{2}x=1-\cos^2x$$$$=\displaystyle \int \dfrac {1-\cos^2\alpha-(1-\cos^2x)}{\cos x-\cos \alpha}$$$$=\displaystyle \int \dfrac {\cos^2 x-\cos^2\alpha}{\cos x-\cos \alpha}\cdot dx$$$$=\displaystyle \int (\cos x+\cos \alpha)\cdot dx$$$$=\sin x+(\cos\alpha)\cdot x+c.$$So, $$f(x)=\sin x$$ and $$A=\cos\alpha.$$Mathematics

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