int frac{sin x + csc x}{tan x} , dx = | vedclass

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Question

(A) We are given the integral $I = \int \frac{\sin x + \csc x}{	an x} \, dx$. First,simplify the integrand: $\frac{\sin x + \csc x}{	an x} = \frac{\

Vedclass · Accepted Answer

$\sin x - \csc x + c$

Vedclass · Answer

(A) We are given the integral $I = \int \frac{\sin x + \csc x}{	an x} \, dx$. First,simplify the integrand: $\frac{\sin x + \csc x}{	an x} = \frac{\sin x}{	an x} + \frac{\csc x}{	an x}$ $= \frac{\sin x}{(\sin x / \cos x)} + \frac{(1 / \sin x)}{(\sin x / \cos x)}$ $= \cos x + \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = \cos x + \csc x \cot x$. Now,integrate term by term: $I = \int (\cos x + \csc x \cot x) \, dx$ $= \int \cos x \, dx + \int \csc x \cot x \, dx$ $= \sin x - \csc x + c$.

$\int \frac{\sin x + \csc x}{\tan x} \, dx = $

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