How do you find the integral of $\int \frac{1}{1 + tan (x)}$ $int 1/(1 + tan(x))$ ?

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How do you find the integral of $\int \frac{1}{1 + tan (x)}$ $int 1/(1 + tan(x))$ ?

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Answer 1 · 2018-04-02T16:54:44

$\int \frac{cos x}{cos + sin x} d x = \frac{1}{2} (x + ln (cos x + sin x)) + c$ $intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c$

Explanation:

Let

$I = \int \frac{1}{1 + tan x} d x$ $I=int1/(1+tanx)dx$

$I = \int \frac{cos x}{cos + sin x} d x$ $I=intcosx/(cos+sinx)dx$

$cos x = l (cos x + sin x) + m (- sin x + cos x) + n$ $cosx=l(cosx+sinx)+m(-sinx+cosx)+n$

$cos x = (l + m) cos x + (l - m) sin x + n$ $cosx=(l+m)cosx+(l-m)sinx+n$

Equating the coefficients of cosx sinx and constants

$l + m = 1, l - m = 0, n = 0$ $l+m=1, l-m=0, n=0$

$l = \frac{1}{2}, n = \frac{1}{2}$ $l=1/2, n=1/2$

$cos x = \frac{1}{2} (cos x + sin x) + \frac{1}{2} (- sin x + cos x) + 0$ $cosx=1/2(cosx+sinx)+1/2(-sinx+cosx)+0$

$I = \int \frac{cos x}{cos + sin x} d x$ $I=intcosx/(cos+sinx)dx$

$I = \frac{1}{2} \int \frac{cos x + sin x}{cos x + sin x} d x + \frac{1}{2} \int \frac{- sin x + cos x}{cos x + sin x} d x$ $I=1/2int(cosx+sinx)/(cosx+sinx)dx+1/2int(-sinx+cosx)/(cosx+sinx)dx$

$I = \frac{1}{2} (I_{1} + I_{2})$ $I=1/2(I_1+I_2)$

where,

$I_{1} = \int \frac{cos x + sin x}{cos x + sin x} d x = \int 1 d x = x$ $I_1=int(cosx+sinx)/(cosx+sinx)dx=int1dx=x$

$I_{1} = x$ $I_1=x$

$I_{2} = \int \frac{- sin x + cos x}{cos x + sin x} d x$ $I_2=int(-sinx+cosx)/(cosx+sinx)dx$

let

$t = cos x + sin x$ $t=cosx+sinx$

$\frac{d t}{d x} = - sin x + cos x$ $dt/dx=-sinx+cosx$

$d t = (- sin x + cos x) d x$ $dt=(-sinx+cosx)dx$

$\int \frac{- sin x + cos x}{cos x + sin x} d x = \int \frac{(- sin x + cos x) d x}{cos x + sin x} = \int \frac{d t}{t} = ln t$ $int(-sinx+cosx)/(cosx+sinx)dx=int((-sinx+cosx)dx)/(cosx+sinx)=intdt/t=lnt$

$I_{2} = ln (cos x + sin x)$ $I_2=ln(cosx+sinx)$

$I = \frac{1}{2} (I_{1} + I_{2})$ $I=1/2(I_1+I_2)$

$\int \frac{cos x}{cos + sin x} d x = \frac{1}{2} (x + ln (cos x + sin x)) + c$ $intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c$

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