Find the indefinite integral $\int \frac{dx}{1-\cot x}$. If the integral evaluates to $Ax + B \ln|\s

Question

Find the indefinite integral $\int \frac{dx}{1-\cot x}$. If the integral evaluates to $Ax + B \ln|\sin x - \cos x| + C$, find the value of $10(A+B)$.

Professor PC Thomas Classes · Answer

Rewrite the integrand: $\frac{1}{1 - \cos x/\sin x} = \frac{\sin x}{\sin x - \cos x}$. Express the numerator as a linear combination of the denominator and its derivative: $\sin x = \frac{1}{2}(\sin x - \cos x) + \frac{1}{2}(\cos x + \sin x)$. The integral splits into $\frac{1}{2}\int 1 dx + \frac{1}{2}\int \frac{\cos x + \sin x}{\sin x - \cos x} dx$. This evaluates to $\frac{1}{2}x + \frac{1}{2}\ln|\sin x - \cos x| + C$. Thus $A = 1/2$ and $B = 1/2$. $10(A+B) = 10(1) = 10$.