What is the antiderivative of $\frac{1}{2 + sin x}$ $1/(2+sinx)$ ?

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Answer 1 · 2018-06-07T12:18:42

$I = \frac{2}{\sqrt{3}} {tan}^{- 1} (\frac{2 tan (\frac{x}{2}) + 1}{\sqrt{3}}) + c$ $I=2/sqrt3tan^-1((2tan(x/2)+1)/sqrt3)+c$

$I = \int \frac{1}{2 + sin x} d x$ $I=int1/(2+sinx)dx$

Subst. $tan (\frac{x}{2}) = t \Rightarrow {sec}^{2} (\frac{x}{2}) \cdot \frac{1}{2} d x = d t$ $tan(x/2)=t=>sec^2(x/2)*1/2dx=dt$

$\Rightarrow [1 + {tan}^{2} (\frac{x}{2})] d x = 2 d t \Rightarrow d x = \frac{2 d t}{1 + t^{2}}$ $=>[1+tan^2(x/2)]dx=2dt=>dx=(2dt)/(1+t^2)$

$I = \int \frac{1}{2 + \frac{2 t}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}}$ $I=int1/(2+(2t)/(1+t^2))xx(2dt)/(1+t^2)$

$I = \int \frac{1}{2 + 2 t^{2} + 2 t} \times 2 d t$ $I=int1/(2+2t^2+2t)xx2dt$

$= \int \frac{1}{t^{2} + t + 1} d t$ $=int1/(t^2+t+1)dt$

$= \int \frac{1}{t^{2} + t + \frac{1}{4} + \frac{3}{4}} d t$ $=int1/(t^2+t+1/4 +3/4)dt$

$= \int \frac{1}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d t$ $=int1/((t+1/2)^2+(sqrt3/2)^2)dt$

But, $\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} {tan}^{- 1} (\frac{x}{a}) + c$ $color(red)(int1/(x^2+a^2)dx=1/atan^-1(x/a)+c$

$:.I=1/(sqrt3/2)tan^-1((t+1/2)/(sqrt3/2))+c$

$I=2/sqrt3tan^-1((2t+1)/sqrt3)+c$

Subst. back , $t=tan(x/2)$ ,we get

$I=2/sqrt3tan^-1((2tan(x/2)+1)/sqrt3)+c$

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