What is the antiderivative of #1/(2+sinx)#?

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Answer 1 · 2018-06-07T12:18:42

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

#I=int1/(2+sinx)dx#

Subst. #tan(x/2)=t=>sec^2(x/2)*1/2dx=dt#

#=>[1+tan^2(x/2)]dx=2dt=>dx=(2dt)/(1+t^2)#

#I=int1/(2+(2t)/(1+t^2))xx(2dt)/(1+t^2)#

#I=int1/(2+2t^2+2t)xx2dt#

#=int1/(t^2+t+1)dt#

#=int1/(t^2+t+1/4 +3/4)dt#

#=int1/((t+1/2)^2+(sqrt3/2)^2)dt#

But, #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#