.
Let now continue.
Hi,
Can someone help me with the interval 0 to pie/2 cos to the 5thxdx, please?
This is what I did:
cos2x=1/2(1+cos2x)
int. 0 to pie/2 (cos3x)cos2xdx=
1/2 int. (1+cos2x)cos3dx=
Then would I do a u substitution?
u=cos2x
du=-sin2x(2)dx
Thank you
Hi,
Can you explain to me how you got this, please?.
I understand how you went to the second step, but I don't understand how you got the third equation.
After that I need to do u substitution, right?
t=sinx
dt=cosxdx
int. 1 to pie/1 (1-sin2x) squared cosx
=integ. -sin2xcosx
Now, I just need to plug in pie/2 and 1, right?
Thank you
recall that
therefore,
yesAfter that I need to do u substitution, right?
t=sinx
dt=cosxdx
what are you doing here? if you made the t-substitution, why do you still have x's in the expressions??int. 1 to pie/1 (1-sin2x) squared cosx
=integ. -sin2xcosx
Now, I just need to plug in pie/2 and 1, right?
Thank you
no integration by parts necessary (i see that you have a problem knowing when to use by parts and when to use regular substitution. we must address this).
anyway, as TPH showed us:
We proceed by substitution.
Let
So now, we look on our integral. wherever we see , we replace it with . if we see , we replace it with
So our integral becomes:
Hopefully, you can take it from here
by the way. please use "^" when you want to show that something is a power. so to type you should type cos^5(x) or (cos(x))^5 etc.