Question

Integrals,

Can you please tell me where am I wrong in this one ? http://screencast.com/t/UoHQfYfNfk3

Thanks

Answer 1

1/12?

Answer 2

only in sin term there will be 1/12 as factor and factor in linear term its 1/2

Answer 3

let u = 3x then du/3 =dx
yours is \(\int \dfrac{1}{3}cos^2u du= \dfrac{1}{3}\int cos^2udu\)
now cos^2 u = \(\dfrac{1+cos2u}{2}\)
so, it is \(\dfrac{1}{6}\int (1+cos2u)du\)
=\(\dfrac{1}{6}u +\dfrac{1}{6}\dfrac{sin2u}{2}+C\)
replace u = 3x
=\(\dfrac{x}{2}+\dfrac{sin6x}{12}+C\)

Answer 4

@phi correct me, please

Answer 5

looks good

Answer 6

thank you

Answer 7

thanks