How do you integrate #(sin(2x))/(5-sin(x))^(1/2) dx#?

1 Answer
Apr 9, 2015

First rewrite : #2int (cos(x)sin(x))/sqrt((5-sin(x))) dx #

Then #u = sin(x)# so #du = cos(x) dx#

Now you have #2int u/sqrt(5-u) du#

now substitute again #t = 5-u # so #dt = -du#

and # u = 5-t#

Now it's more easy : #-2int (5-t)/sqrt(t) dt#

#= -2int 5/sqrt(t)-sqrt(t) dt #

#= 2intsqrt(t) dt - 10int1/sqrt(t) dt#

#= 4/3*t^(3/2) - 20sqrt(t) + C#

Substitute back for #t = 5 - u#

#=4/3(5-u)^(3/2)-20sqrt(5-u) + C#

Again for #u = sin(x)#

#=4/3(5-sin(x))^(3/2)-20sqrt(5-sin(x)) + C#

You can factorize if you want