Find the following integral:int frac{1}{1+tan | vedclass

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(A) Let $I = \int \frac{1}{1+	an x} d x$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int \frac{\cos x}{\cos x + \sin x} d x$.Multiply the

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$\frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C$

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(A) Let $I = \int \frac{1}{1+	an x} d x$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int \frac{\cos x}{\cos x + \sin x} d x$.Multiply the numerator and denominator by $2$ and adjust the numerator:$I = \frac{1}{2} \int \frac{2\cos x}{\cos x + \sin x} d x = \frac{1}{2} \int \frac{(\cos x + \sin x) + (\cos x - \sin x)}{\cos x + \sin x} d x$.Split the integral:$I = \frac{1}{2} \int 1 d x + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} d x$.For the second integral,let $u = \cos x + \sin x$,then $du = (-\sin x + \cos x) d x$.Thus,$I = \frac{x}{2} + \frac{1}{2} \int \frac{1}{u} du = \frac{x}{2} + \frac{1}{2} \log |u| + C$.Substituting back $u = \cos x + \sin x$,we get:$I = \frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C$.

Find the following integral:
$\int \frac{1}{1+\tan x} d x$

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Find the following integral:$\int \frac{1}{1+\tan x} d x$