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anonymous
 5 years ago
What is a strategy to evaluate INT(sqrt(ax^2)dx) ?
anonymous
 5 years ago
What is a strategy to evaluate INT(sqrt(ax^2)dx) ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm the least experienced with trig substitution, can you show me the next step? my problem is that i keep ending up with two terms under the sqrt and they don't make another trig identity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x= a sin(t) dx/dt= a cost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i imagine what i'm trying to get is a trig function squared under the square root so that they cancel, is there another way that would get rid of that square root?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey in ur question is it a or a^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...do integration by parts..u know how to do that.......I've done it on paper ..it works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so x=sin (t), and int by parts as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no just integration by parts without any substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so u=ax^2 and dv=sqrt(u)du ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont substitute, take the whole square root thing as first term and 1 as second term. okay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i don't get it, it just seems to prolong the problem. i still have two terms under the sqrt.
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