Greetings and happy holidays
Im recalling a bit of integrals and find this one that I dont see right.
http://img160.imagevenue.com/img.php?ima…
I mean, the first part is well done because I think it uses a reduction formula but the point its where the 8 pop out?
isnt this one evaluated as the int of cos 8tdt? > sin8t.
Thanks in advance.
Take care
Im recalling a bit of integrals and find this one that I dont see right.
http://img160.imagevenue.com/img.php?ima…
I mean, the first part is well done because I think it uses a reduction formula but the point its where the 8 pop out?
isnt this one evaluated as the int of cos 8tdt? > sin8t.
Thanks in advance.
Take care
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u-substitution will show you where that "1/8" came from.
Let u = 8t. Then you have du = 8 dt, or dt = 1/8 du
Then you get cos(8t) dt = cos(u) * 1/8 du.
You can also think of this backwards. If you were doing the derivative of sin(8t), you would need to use the chain rule. First, you do the derivative of the outside, giving you cos(8t), but you'd need to multiply by the derivative of the inside, which is 8. Thus, you get 8cos(8t), but you only want cos(8t), so you'd need to divide by 8 in order to get rid of that 8 that "pops out."
Let u = 8t. Then you have du = 8 dt, or dt = 1/8 du
Then you get cos(8t) dt = cos(u) * 1/8 du.
You can also think of this backwards. If you were doing the derivative of sin(8t), you would need to use the chain rule. First, you do the derivative of the outside, giving you cos(8t), but you'd need to multiply by the derivative of the inside, which is 8. Thus, you get 8cos(8t), but you only want cos(8t), so you'd need to divide by 8 in order to get rid of that 8 that "pops out."
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The integral is correct. Differentiate sin(8t) and tell me if you get cos(8t) * dt or 8 * cos(8t) * dt