New to Integration (and new to forum)

**Krizalid** · September 9th 2009, 07:30 PM

ahhh well, then your substitution will work and $2t\,dt=-2x\,dx\implies t\,dt=-x\,dx.$

**PandaNomium** · September 9th 2009, 07:37 PM

0_0.... it's so simple. -sqrt(9-x^2) +C

thanks so much for your help. do you mind if i ask another question?

i have to have the int of sec (x/2)
am I on the right track?

= 1/cos(x/2)
t=cos(x/2)
t dt= -1/2sin(x/2)

**Krizalid** · September 9th 2009, 07:42 PM

no, that substitution won't work.

that integral is tricky, first put $t=\frac x2$ so that you only care about integrating $\sec t.$

we have $\int{\sec t\,dt}=\int{\frac{\sec ^{2}t+\sec t\tan t}{\sec t+\tan t}\,dt},$ and then put $u=\sec t+\tan t.$

**PandaNomium** · September 9th 2009, 07:55 PM

you might have to walk me through that one. I understand t dt= sec(t)tan(t), where did the rest of that come from?

Thread: New to Integration (and new to forum)

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