How do you find the antiderivative of $\frac{sin x + cos x}{tan x}$ $(sinx + cosx)/tanx$ ?

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How do you find the antiderivative of $\frac{sin x + cos x}{tan x}$ $(sinx + cosx)/tanx$ ?

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Answer 1 · 2016-07-27T11:20:54

$sin x + ln ∣ ∣ tan (\frac{x}{2}) ∣ ∣ + cos x + C$ $sinx+ln|tan(x/2)|+cosx+C$

Explanation:

$I = \int \frac{sin x + cos x}{tan x} d x = \int (sin x + cos x) \frac{cos x}{sin x} d x$ $I=int(sinx+cosx)/tanxdx=int(sinx+cosx)cosx/sinxdx$

$= \int cos x d x + \int \frac{{cos}^{2} x}{sin x} d x = sin x + \int \frac{1 - {sin}^{2} x}{sin x} d x$ $=intcosxdx+intcos^2x/sinxdx=sinx+int(1-sin^2x)/sinxdx$

$= sin x + \int {\frac{1}{sin x} - sin x} d x$ $=sinx+int{1/sinx-sinx}dx$

$= sin x + \int (csc x - sin x) d x$ $=sinx+int(cscx-sinx)dx$

$:. I=sinx+ln|tan(x/2)|+cosx+C$

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How do you find the antiderivative of $\frac{sin x + cos x}{tan x}$ $(sinx + cosx)/tanx$ ?

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Explanation:

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How do you find the antiderivative of $\frac{sin x + cos x}{tan x}$ $(sinx + cosx)/tanx$ ?