How do you integrate #int cosx/sqrt(1-2sinx) #?
1 Answer
Jul 30, 2016
Explanation:
We have:
#intcosx/sqrt(1-2sinx)dx#
Let:
#intcosx/sqrt(1-2sinx)dx=-1/2int(-2cosx)/sqrt(1-2sinx)dx#
#=-1/2int(du)/sqrtu=-1/2intu^(-1/2)du=-1/2(u^(-1/2+1)/(-1/2+1))+C#
#=-u^(1/2)+C=-sqrt(1-2sinx)+C#