What is the antiderivative of $\frac{1}{1 + sin x} d x$ $1/ (1+sinx) dx$ ?

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What is the antiderivative of $\frac{1}{1 + sin x} d x$ $1/ (1+sinx) dx$ ?

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Answer 1 · 2018-05-08T07:37:09

$I = tan x - sec x + c$ $I=tanx-secx+c$

Explanation:

We know that,
$(1) {sin}^{2} θ + {cos}^{2} θ = 1$ $color(red)((1)sin^2theta+ cos^2theta=1$
$(2) \frac{1}{cos θ} = sec θ and \frac{sin θ}{cos θ} = tan θ$ $color(blue)((2)1/costheta=sectheta and sintheta/costheta=tantheta$
$(3) \int {sec}^{2} x d x = tan x + c$ $color(violet)((3)intsec^2xdx=tanx+c$
$(4) \int sec x tan x d x = sec x + c$ $color(violet)((4)intsecxtanxdx=secx+c$
Here,

$I = \int \frac{1}{1 + sin x} d x = \int \frac{(1 - sin x)}{(1 + sin x) (1 - sin x)} d x$ $I=int1/(1+sinx)dx=int((1-sinx))/((1+sinx)(1-sinx))dx$

$\Rightarrow I = \int \frac{1 - sin x}{1 - {sin}^{2} x} d x$ $=>I=int(1-sinx)/(1-sin^2x)dx$

$=int(1-sinx)/cos^2xdx...tocolor(red)(Apply(1)$

$=int[1/cos^2x-sinx/cos^2x]dx$

$=int[sec^2x-secxtanx]dx...tocolor(blue)(Apply(2)$

Using , $color(violet)((3) and (4)$ , we get

$I=tanx-secx+c$

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What is the antiderivative of $\frac{1}{1 + sin x} d x$ $1/ (1+sinx) dx$ ?

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