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, 02:00 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, 02:27 PMSorobanRe: Dose this integration seem right
Hello, duggielanger!

Quote:

I used the double angle formulas and got: .

Is this right? . no

If you differentiate your answer, you doget*not*

. . . . . . . . . .

. . . . . . . . . .

- Feb 13th 2013, 02:43 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