# Missing steps in book

• May 22nd 2010, 09:25 AM
calypso
Missing steps in book
I am trying to work out how to get from the integral below to the answer just below it.

$N = \frac{yR^2}{sin^2x} \int{(cosx-1)cosxsinx dx}$

with limits of pie and x

$N = \frac{yR^2}{6} \frac{5(1-cosx) + 2cos^2x}{1-cosx}$

After using the following intergrals

$\int{cosxsinx dx} = \frac{sin^2x}{2}$

and

$\int{cos^2xsinx dx} = \frac{-cos^3x}{3}$

I get

$N = \frac{yR^2}{6sin^2x} [3sin^2x + 2cos^2x + 2]$

however

1. I cannot rearange it to get the correct answer
2. If i substitute a value such as x = 180, my answer and the book's answer do not match?

Calypso
• May 22nd 2010, 11:57 AM
Quote:

Originally Posted by calypso
I am trying to work out how to get from the integral below to the answer just below it.

$N = \frac{yR^2}{sin^2x} \int{(cosx-1)cosxsinx dx}$

with limits of pie and x

$N = \frac{yR^2}{6} \frac{5(1-cosx) + 2cos^2x}{1-cosx}$

After using the following intergrals

$\int{cosxsinx dx} = \frac{sin^2x}{2}$

and

$\int{cos^2xsinx dx} = \frac{-cos^3x}{3}$

I get

$N = \frac{yR^2}{6sin^2x} [3sin^2x + 2cos^2x + 2]$

however

1. I cannot rearange it to get the correct answer
2. If i substitute a value such as x = 180, my answer and the book's answer do not match?

Calypso

Hi Calypso,

one of your integrals isn't evaluated conveniently..

$\int{-cosxsinx}dx$

$u=-cosx,\ \frac{du}{dx}=sinx,\ du=sinxdx$

$\int{u}du=\frac{u^2}{2}=\frac{cos^2x}{2}$

Then you get

$\frac{yR^2}{sin^2x}\left(\int{cos^2xsinx}dx-\int{cosxsinx}dx\right)$

$=\frac{yR^2}{sin^2x}\left[-\frac{1}{3}cos^3x+\frac{1}{2}cos^2x\right]$ from x to ${\pi}$

$=\frac{yR^2}{sin^2x}\left[\frac{1}{3}cos^3x-\frac{1}{2}cos^2x\right]$ from ${\pi}$ to x

This gives

$\frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}-\frac{cos^3{\pi}}{3}+\frac{cos^2{\pi}}{2}\right)$

$=\frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}-\frac{(-1)^3}{3}+\frac{(-1)^2}{2}\right)$

$=\frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}+\frac{1}{3}+\frac{1}{2}\right)$

$=\frac{yR^2}{6sin^2x}\left(2cos^3x-3cos^2x+5\right)$

$=\frac{yR^2}{6}\left(\frac{2cos^3x+2cos^2x-5cos^2x+5}{1-cos^2x}\right)$

$=\frac{yR^2}{6}\left(\frac{2cos^2x(1+cosx)+5(1-cos^2x)}{(1-cosx)(1+cosx)}\right)$

$=\frac{yR^2}{6}\left(\frac{2cos^2x(1+cosx)+5(1-cosx)(1+cosx)}{(1-cosx)(1+cosx)}\right)$

$=\frac{yR^2}{6}\left(\frac{2cos^2x+5(1-cosx)}{1-cosx}\right)$
• May 22nd 2010, 12:41 PM
calypso
Thanks Archie, I have been trying to do that all afternoon