If int frac{1}{1-cot x} dx = Ax + B log |sin | vedclass

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(A) Let $I = \int \frac{1}{1-\cot x} dx$. Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\

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(A) Let $I = \int \frac{1}{1-\cot x} dx$. Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\sin x}{\sin x - \cos x} dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} dx$. Rewrite the numerator: $2 \sin x = (\sin x - \cos x) + (\sin x + \cos x)$. Thus,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) dx$. Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) dx$. $I = \frac{1}{2} \left( \int 1 dx + \int \frac{1}{u} du \right) = \frac{1}{2} (x + \log |u|) + C = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C$. Comparing with $Ax + B \log |\sin x - \cos x| + C$,we get $A = \frac{1}{2}$ and $B = \frac{1}{2}$. Therefore,$A + B = \frac{1}{2} + \frac{1}{2} = 1$.

If $\int \frac{1}{1-\cot x} dx = Ax + B \log |\sin x - \cos x| + C$,then $A + B = \dots$

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