# Dose this integration seem right

• Feb 13th 2013, 02:00 PM
duggielanger
Dose 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 PM
Soroban
Re: Dose this integration seem right
Hello, duggielanger!

Quote:

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

I used the double angle formulas and got: . $\tfrac{1}{32}\sin(16u)+\tfrac{1}{2}u+C$
Is this right? . no

If you differentiate your answer, you do not get $\sin^2(8u)$

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

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

. . . . . . . . . . $=\;\tfrac{1}{2}u - \tfrac{1}{32}\sin(16u) + C$
• Feb 13th 2013, 02:43 PM
duggielanger
Re: 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