Sorry, you're right - parts wouldn't work as you'll end up with a cycle of trig functions I edited it, you're supposed to use the equations as above stated by KingAaran.

Reply 6

crashMATHS

Original post by mil88

Ok thanks, btw are they on the formula sheet (how did you find out about them?)

They are on the formula sheet :smile:

It is usually quite helpful in such cases, because you can split the integral into one involving sin(ax) and one involving cos(bx) which makes it relatively simplistic after!

Reply 7

MathsAndChess

Parts would work:

Unparseable latex formula:

[br]\displaystyle[br]\begin{align*} I &= \int \sin6x\cos9xdx\\&= g(x) - a\int\cos6x\sin9xdx\\&= g(x) - a\left( h(x) - b\int \sin6x\cos9xdx\right )\\&= g(x) - a\left( h(x) - bI\right)\\\end{align*}[br]

Solve for

I

(edited 8 years ago)

Reply 8

username2097061

Original post by kingaaran

They are on the formula sheet :smile:

It is usually quite helpful in such cases, because you can split the integral into one involving sin(ax) and one involving cos(bx) which makes it relatively simplistic after!

Indeed.

Reply 9

Zacken

Original post by MathsAndChess

Parts would work:
...

But very overkill! :tongue:

Reply 10

username2097061

Original post by MathsAndChess

Parts would work:

Unparseable latex formula:

[br]\displaystyle[center]\begin{align*} I &= \int \sin6x\cos9xdx\\&= g(x) - a\int\cos6x\sin9xdx\\&= g(x) - a\left( h(x) - b\int \sin6x\cos9xdx\right )\\&= g(x) - a\left( h(x) - bI\right)\\\end{align*}[/center]

Ok thanks!

Reply 11

crashMATHS