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by **doggy** on Fri Apr 17, 2009 12:33 am

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)?

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by **Martingale** on Fri Apr 17, 2009 12:40 am

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)?

$\int (1-cos(\frac{t}{6}))^6 sin(\frac{t}{6})dt$

how about $u=1-cos(\frac{t}{6})$

by **doggy** on Fri Apr 17, 2009 1:51 am

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

by **Martingale** on Fri Apr 17, 2009 2:01 am

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:
$\int (1-cos(\frac{t}{6}))^6 sin(\frac{t}{6})dt$

how about $u=1-cos(\frac{t}{6})$

$du=sin(\frac{t}{6})\cdot\frac{1}{6}dt$

so

$6du=sin(\frac{t}{6})dt$

$\int (1-cos(\frac{t}{6}))^6 sin(\frac{t}{6})dt=6\int (u)^6du=\cdots$

Integrals: indefinite integral (1-cos(t/6))^6 sin(t/6) dt