Simple Calculus (indefinite integral)

AryRezvani
Messages
65
Reaction score
0

Homework Statement



Find the indefinite integral:
∫ Cos32x Sin22x dx

Homework Equations



None required

The Attempt at a Solution



Lost on where to start. If someone could just start me off.
 
Physics news on Phys.org
either use the identity
cos3(2x)sin2(2x)=(1/16) (2 cos(2 x)-cos(6 x)-cos(10 x))
or
cos3(2x)sin2(2x)=(1/2)(sin2(2 x)-sin4(2 x))(sin(2x))'
and the substitution u=sin(2x)
 
AryRezvani said:

Homework Statement



Find the indefinite integral:
∫ Cos32x Sin22x dx

Homework Equations



None required

The Attempt at a Solution



Lost on where to start. If someone could just start me off.
Write cos3(2x) as cos(2x) cos2(2x)

Then use the identity,
cos2(2x)=1-sin2(2x)​
After that, a simple substitution should work fine.
 
SammyS said:
Write cos3(2x) as cos(2x) cos2(2x)

Then use the identity,
cos2(2x)=1-sin2(2x)​
After that, a simple substitution should work fine.

Knocked it out, thanks guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top