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

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- Sep 9th 2009, 07:30 PM #16

- Sep 9th 2009, 07:37 PM #17

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- Sep 9th 2009, 07:42 PM #18
no, that substitution won't work.

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

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

- Sep 9th 2009, 07:55 PM #19

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