To integrate 1/sq rt(1-v^2) with respect to v, we need to use trigo substitution to integrate rite?(adsbygoogle = window.adsbygoogle || []).push({});

Why we cant do it this way:

1/sq rt(1-v^2) = (1-v^2)^(-1/2)

Then integrate and it becomes (-1/3v)(1-v^2)^(3/2) ??

And how do we integrate sq rt[1+(2y)^2] with respect to y?

Can someone help me pls? Thanks!!!

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# I need help with Integration pls

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