int frac{dx}{tan x + cot x} = | vedclass

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(D) We have the integral $I = \int \frac{dx}{	an x + \cot x}$.First,express $	an x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:$	an x + \cot x

Vedclass · Accepted Answer

$-\frac{\cos 2x}{4} + c$

Vedclass · Answer

(D) We have the integral $I = \int \frac{dx}{	an x + \cot x}$.First,express $	an x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:$	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$.Substituting this into the integral:$I = \int \sin x \cos x \, dx$.Multiply and divide by $2$:$I = \frac{1}{2} \int 2 \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx$.Integrating $\sin 2x$ gives $-\frac{\cos 2x}{2}$:$I = \frac{1}{2} \left( -\frac{\cos 2x}{2} ight) + c = -\frac{\cos 2x}{4} + c$.

$\int \frac{dx}{\tan x + \cot x} = $

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