Calculus II - Trigonometric Integrals

Homework Statement

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

Homework 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)

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

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eumyang
Homework Helper
sin^2 x = ( 1 - cos 2x )/2
cos^2 x = ( 1 - sin 2x )/2
The bolded is not right. It should be
$$\cos^2 x = \frac{1 + \cos 2x}{2}$$

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:

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this

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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

oh man im an idiot thanks

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

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

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

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

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...

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

anyone?

Ray Vickson
Homework Helper
Dearly Missed
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...

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
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

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

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...