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