- #1

MillerL7

- 14

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I think u=sin^3

so du/dx=cos^3

du=cos^3dx

but then I am really not sure if that is correct, the trig functions confuse me a bit. Please help me, thank you!

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- Thread starter MillerL7
- Start date

- #1

MillerL7

- 14

- 0

I think u=sin^3

so du/dx=cos^3

du=cos^3dx

but then I am really not sure if that is correct, the trig functions confuse me a bit. Please help me, thank you!

- #2

RyanSchw

- 36

- 0

Since this is in the form of [itex] Sin^n(u)Cos^m(u) [/itex] where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?

- #3

Dick

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- #4

Dick

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Since this is in the form of [itex] Sin^n(u)Cos^m(u) [/itex] where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?

Why do you say trigonometric integrals have special rules? AFAIK they have the same rules as other integrals. What's special?

- #5

RyanSchw

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- #6

Dick

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Right. Sorry, I just jumped on you when you said you can't use a u substitution. Because the answer IS a u substitution. Together with a specific trig rule.

Last edited:

- #7

silver-rose

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You need to reduce the power of the sin from [tex] \sin{^3 (13x)} [/tex] into [tex] \sin{13x} [/tex]

- #8

HallsofIvy

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- #9

MillerL7

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I tried that formula and got it incorrect...

- #10

Dick

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I tried that formula and got it incorrect...

That's not very helpful. How are we supposed to know how you got it wrong?

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