polar integral

**kittycat** · August 1st 2007, 09:22 AM

8/3 times the integral (from 0 to pi) of sin^3 theta d(theta).

Thank you very much.

**galactus** · August 1st 2007, 09:31 AM

$\int{sin^{n}(x)}dx=\frac{-1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\int{sin^{n-2}(x)}dx$

You can use the formula to check if you wish.

$\int{sin^{3}(x)}dx$

$\int{sin(x)sin^{2}(x)}dx$

$\int{sin(x)(1-cos^{2}(x))}dx$

Let $u=cos(x), \;\ du=-sin(x)dx$

$\int{(u^{2}-1)}du=\frac{u^{3}}{3}-u$

Resub:

$\frac{cos^{3}}{3}-cos(x)+C$

or by the reduction formula:

$\frac{-sin^{2}(x)cos(x)}{3}-\frac{2}{3}cos(x)+C$

Same thing.

**tukeywilliams** · August 1st 2007, 09:52 AM

$\frac{8}{3} \int_{0}^{\pi} \sin^{3} \theta \ d \theta = \frac{8}{3} \int_{0}^{\pi} (1- \cos^{2} \theta) \sin \theta \ d \theta$ . Let $u = \cos \theta$ , then $du = - \sin \theta \ d \theta$ or

$-\frac{8}{3} \int_{0}^{\pi} 1-u^2 \; du = u - \frac{u^3}{3} = -\frac{8}{3} \left[\cos \theta - \frac{\cos^{3} \theta}{3} \right ]$ .

What would it be from $0$ to $\pi$ ? $\frac{32}{9}$ .

**kittycat** · August 1st 2007, 10:13 AM

Thank you very much again turkeywilliam. Yes, the answer is 32/9. Thanks!

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