int(cot x cot (x+alpha)+1) d x= | vedclass

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Question

(A) We have the integral $I = \int(\cot x \cot (x+\alpha)+1) d x$. Using the identity $\cot A \cot B + 1 = \frac{\cos A \cos B + \sin A \sin B}{\sin A

Vedclass · Accepted Answer

$\cot \alpha \log \left|\frac{\sin x}{\sin (x+\alpha)}\right|+c$

Vedclass · Answer

(A) We have the integral $I = \int(\cot x \cot (x+\alpha)+1) d x$. Using the identity $\cot A \cot B + 1 = \frac{\cos A \cos B + \sin A \sin B}{\sin A \sin B} = \frac{\cos(A-B)}{\sin A \sin B}$, we get: $I = \int \frac{\cos(x - (x+\alpha))}{\sin x \sin (x+\alpha)} d x = \int \frac{\cos(-\alpha)}{\sin x \sin (x+\alpha)} d x = \cos \alpha \int \frac{1}{\sin x \sin (x+\alpha)} d x$. Multiply and divide by $\sin \alpha$: $I = \frac{\cos \alpha}{\sin \alpha} \int \frac{\sin \alpha}{\sin x \sin (x+\alpha)} d x = \cot \alpha \int \frac{\sin((x+\alpha)-x)}{\sin x \sin (x+\alpha)} d x$. Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $I = \cot \alpha \int \frac{\sin(x+\alpha) \cos x - \cos(x+\alpha) \sin x}{\sin x \sin (x+\alpha)} d x$. $I = \cot \alpha \int (\cot x - \cot (x+\alpha)) d x$. Integrating, we get: $I = \cot \alpha (\log |\sin x| - \log |\sin (x+\alpha)|) + c = \cot \alpha \log \left|\frac{\sin x}{\sin (x+\alpha)}\right| + c$.

$\int(\cot x \cot (x+\alpha)+1) d x=$

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