How do I evaluate the indefinite integral (1-cos(t/6))^6 sin(t/6)?

Would I use a u substitution of u= sin(t/6)?

How do I evaluate the indefinite integral (1-cos(t/6))^6 sin(t/6)?

Would I use a u substitution of u= sin(t/6)?

Would I use a u substitution of u= sin(t/6)?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote:How do I evaluate the indefinite integral (1-cos(t/6))^6 sin(t/6)?

Would I use a u substitution of u= sin(t/6)?

how about

So then the answer would be (u^7/7)(6u) = ((1-cos(t/6))^7/7)(6(1-cos(t/6))

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote:So then the answer would be (u^7/7)(6u) = ((1-cos(t/6))^7/7)(6(1-cos(t/6))

Martingale wrote:

how about

so