# Homework Help: Calculus II - Trigonometric Integrals

1. Jul 30, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

Apparently I'm doing something wrong. I'm kind of lost as to what because I looked over my work several times.

2. Relevant equations

sin^2 x = ( 1 - cos 2x )/2
cos^2 x = ( 1 - sin 2x )/2
integral sin(x) dx = -cos(x)
integral cos(x) dx = sin(x)
3. The attempt at a solution

SEE ATTACHMENT
Wolfram Alpha says the correct solution is 1/64 (24 x+8 sin(4 x)+sin(8 x))+constant
Thanks for any help you can provide!

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Last edited: Jul 30, 2011
2. Jul 30, 2011

### eumyang

The bolded is not right. It should be
$$\cos^2 x = \frac{1 + \cos 2x}{2}$$

3. Jul 30, 2011

### GreenPrint

Thanks for your response. My book specifically told me those were true and even use the fact that they were to evaluate a integral. For example:

Last edited: Jul 30, 2011
4. Jul 30, 2011

### GreenPrint

this

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5. Jul 30, 2011

### GreenPrint

So i guess there's a plus sign and I forgot to include it becuase i used minus instead

sin^2 x = ( 1 - cos 2x )/2
cos^2 x = ( 1 + sin 2x )/2

cos^2 x = ( 1 + sin 2x )/2 instead of cos^2 x = ( 1 - sin 2x )/2

That is one mistake in my work... assuming that formula is right

6. Jul 30, 2011

### GreenPrint

oh man im an idiot thanks

7. Jul 30, 2011

### GreenPrint

I guess I'm still doing something wrong, don't know what though

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8. Jul 30, 2011

### JHamm

cos^2(x) = 1/2(1 + cos(2x))

9. Jul 30, 2011

### GreenPrint

Ya I know. I still don't see what I did wrong though in post #7

10. Jul 30, 2011

### eumyang

It's actually the same as the answer you got from Wolfram. You just need to factor out another 1/8.
$\frac{1}{8}\left( 3x + \sin 4x + \frac{\sin 8x}{8}\right) + C$
$= \frac{1}{8}\left( \frac{24x}{8} + \frac{8\sin 4x}{8} + \frac{\sin 8x}{8}\right) + C$
Do you see it now?

11. Jul 30, 2011

### GreenPrint

The answers don't seem to match though for some reason and I don't have a clue as to what I'm doing wrong...

The answer I got was
1/8 [ 3x + sin(4x) + sin(8x)/8 ]+ c
wolfram alpha returned
1/32 (12 x-8 sin(2 x)+sin(4 x))+constant

if they are equal to each other I should be able to plug in a value for x in both expressions and get the same results... this however doesn't occur

1/8 [ 3*5 + sin(4*5) + sin(8*5)/8 ]+ c is about 2.000760549 + c

1/32 (12*5 -8 sin(2*5)+sin(4*5))+constant is about 2.039534817 + c

It would appear to be that i am off by a little bit... which is strange cause I don't see what I did wrong

and ya they would appear to be equal to each other but i guess they are not

12. Jul 30, 2011

### GreenPrint

anyone?

13. Jul 30, 2011

### Ray Vickson

For *indefinite* integrations, the C is an arbitrary constant. There is no reason the two C's should be the same. However, you ought to get the same results for the definite integral from a to b. Have you tried it?

RGV

14. Jul 30, 2011

### GreenPrint

Ya I got different results when I integrated both functions from 2 to 10

for 1/8 [ 3*x + sin(4*x) + sin(8*x)/8 ]
I got roughly 18.01464011

for 1/32 (12*x -8 sin(2*x)+sin(4*x))
I got roughly 18.13678944

15. Jul 30, 2011

### Bohrok

WA gave me 1/64(24x + 8sin(4x) + sin(8x)) + constant, which is equivalent to the answer you say you got above.
For the other answer that you say WA gave you, I took its derivative in WA and it gave sin4x, so it looks like you had a mistake when you entered it in.

16. Jul 30, 2011

### GreenPrint

huh weird thanks