The question h(u)=sin^2(8u)

I used the double angle formulas and the answer I got is

1/32sin(16u)+1/2u+c

Is this right

Printable View

- Feb 13th 2013, 01:00 PMduggielangerDose this integration seem right
The question h(u)=sin^2(8u)

I used the double angle formulas and the answer I got is

1/32sin(16u)+1/2u+c

Is this right - Feb 13th 2013, 01:27 PMSorobanRe: Dose this integration seem right
Hello, duggielanger!

Quote:

$\displaystyle \int \sin^2(8u)\,du$

I used the double angle formulas and got: .$\displaystyle \tfrac{1}{32}\sin(16u)+\tfrac{1}{2}u+C$

Is this right? . no

If you differentiate your answer, you doget $\displaystyle \sin^2(8u)$*not*

$\displaystyle \int\sin^2(8y)\,du \;=\;\tfrac{1}{2}\int\big[1 - \cos(16u)\big]\,du$

. . . . . . . . . . $\displaystyle =\;\tfrac{1}{2}\left[u - \tfrac{1}{16}\sin(16u)\right] + C $

. . . . . . . . . . $\displaystyle =\;\tfrac{1}{2}u - \tfrac{1}{32}\sin(16u) + C$

- Feb 13th 2013, 01:43 PMduggielangerRe: Dose this integration seem right
Ok i can't see where I've gone wrong in my workings out so I will post them and could you have a look please (can't find a theta sign so i'll type

it)

sin^2(8u)

where

sin^2=1-cos^2theta here theta=8u

so

sin^2(8u)=1-cos^2(8u)

sin^2(8u)=1-cos^2(16u)

sin^2(8u)=1/2(cos(16u)+1)

sin^2(8u)=(1/2cos(16u)+1/2)=1/32sin(16u)-1/2u+c

Thank you