Question

Integral (2-3x) cos9xdx

Answer 1

You can break this up like this:
\[\int2\cos(9x)dx - 3\int x\cos(9x)dx\]

The first integral should be easy to compute, the second will involve integration by parts.
http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

Hope this helps.

Answer 2

tnx

Answer 3

\[\int\limits_{ }^{}\cos9xdx=1/9 \sin9x\]
\[\int\limits_{ }^{}x*\cos 9x dx=\]
u=x du=dx
dv=cos 9x v=1/9 sin9x
(integration by parts)
then:
\[\int\limits_{ }^{}x* \cos9x = 1/9 *x*\sin9x - 1/9 \int\limits_{ }^{}\sin9x dx=\]

can you finish it?

Answer 4

yea thanks, i have done it using integration by part,
u = (2-3x)
dv = cos9xdx

Answer 5

awesome!