Hi everybody !
First of all, I'm new and I apologize, English is not my first language...
And may I require your help ?
So here is the problem :
int[-pi/2, pi/2] (cos^3(t)/(1+sin²(t)) dt
I tried something :
u = sin(t) => t = arcsin (u)
dt = arcsin'(u) du
And we have to change the "frames" of the integral:
For t= -pi/2 => u = -1
For t= pi/2 => u=1
It becomes: int[-1, 1] ((u'^3) du / (1+u²)(sqrt(1-u²)))
And I'm quite stuck ...
I recognise, int(u'/(1+u²)) is arctan(u) but I can't do anything with the rest, plus it's a multiplication, so I tried to integrate the whole thing in parts (int(u'v)= uv - int(v'u)) but it gives me something awful to solve haha
Many thanks for your help, and sorry again if it's not that understandable...
~Manon
First of all, I'm new and I apologize, English is not my first language...
And may I require your help ?
So here is the problem :
int[-pi/2, pi/2] (cos^3(t)/(1+sin²(t)) dt
I tried something :
u = sin(t) => t = arcsin (u)
dt = arcsin'(u) du
And we have to change the "frames" of the integral:
For t= -pi/2 => u = -1
For t= pi/2 => u=1
It becomes: int[-1, 1] ((u'^3) du / (1+u²)(sqrt(1-u²)))
And I'm quite stuck ...
I recognise, int(u'/(1+u²)) is arctan(u) but I can't do anything with the rest, plus it's a multiplication, so I tried to integrate the whole thing in parts (int(u'v)= uv - int(v'u)) but it gives me something awful to solve haha
Many thanks for your help, and sorry again if it's not that understandable...
~Manon
