# The integral of (sin 3t)^5 cos t dt

1. Mar 27, 2005

### PhysicsMajor

Hey folks,

HERE GOES....

the integral of (sin 3t)^5 cos t dt

i believe you have to use u substitution but i am having trouble getting it set up correctly.

Thanks for any input

2. Mar 27, 2005

### whozum

$$\int{sin(3t)^5}cos(t)}{dt}$$

u = sin(3t)
du = 3cos(3t) dt

$$1/3\int{u^5*3cos(3t)}{dt}$$

$$1/3\int{u^5}{du}$$

3. Mar 27, 2005

### Nylex

How does that work? What happened to the cos t?

4. Mar 27, 2005

### whozum

From converting the integral to one with respect to u instead of t. The conversion factor is right above the integral

5. Mar 27, 2005

### whozum

Ugh I did it wrong hold on. Im so tired

6. Mar 27, 2005

### Galileo

There may be easier ways, but you could expand:

$$\sin(3t)=4sin(t)cos(t)^2-sin(t)$$
by exploiting deMoivre's theorem.

7. Mar 27, 2005

### Nylex

Mathworld's Integrator gives a long and not-very-nice answer btw.

8. Mar 27, 2005

### whozum

Then you'd have to raise that to the fifth and foil and blah blah blah. Its a real complicated integral.

9. Mar 27, 2005

### Galileo

Well, you ARE multiplying a power of a sin with a cosine with a different argument, I don't expect the answer to be nice.
Besides, you just have to develop a 'work-attitude' in some situations. Roll up your sleeves and do it. It may be the fastest way. It won't take more than a few minutes, while looking for a possible easier way probably takes longer.

10. Mar 27, 2005

### whozum

$$sin(3t) = sin(2t+t) = sin(2t)cos(t) + sin(t)cos(2t) = 2sin(t)cos^2(t) + sin(t)(cos^2(t)-sin^2(t))$$

$$= 2sin(t)(1-sin^2(t)) + sin(t)(1-2sin^2(t)) = 2sin(t)-2sin^3(t) + sin(t)-2sin^3(t) = 3sin(t)-4sin^3(t)$$

$$\int{sin(3t)^5cos(t)}{dt}$$

$$\int{(3sin(t)-5sin^3(t))^5cos(t)}{dt}$$

$$u=sin(t), du = cos(t) dt$$

$$\int{(3u-4u^3)^5}{du}$$

I think thats as good as it gets if I didnt make any errors.

Last edited: Mar 27, 2005
11. Mar 27, 2005

### dextercioby

This is integral is the typical example of an easy & but messy integral.Eas,because you know what to do to get to the result and messy,because it would take a page of writing to do it...

Daniel.

P.S.$$\sin 3x=-4\sin^{3}x+3\sin x$$

12. Mar 27, 2005

### dextercioby

$$\int \left( \left( \sin 3x\right) ^5\cos x\right) dx= -\frac 1{512}\cos 16x-\frac 1{448}\cos 14x+\frac 1{64}\cos 10x+ \frac 5{256}\cos 8x-\frac 5{64}\cos 4x-\frac 5{32}\cos 2x +C$$

Daniel.

13. Mar 27, 2005

### dextercioby

Making the substitution

$$\sin x=u$$

u'll need to evaluate this cutie pie

$$\int \left(-4u^3+3u\right)^{5} \ du$$

So use the binomial formula.

Daniel.

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