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lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{dy}{dx} = \frac{\sec^2 y}{1 + x^2}\] divide both sides by sec^2 y \[\implies \frac{dy}{\sec^2 y dx} = \frac{1}{1+x^2}\] now multiply both sides by dx \[\implies \frac{dy}{\sec^2 y} = \frac{dx}{1+x^2}\] now you can integrate both sides \[\implies \int \frac{dy}{\sec^2 y} = \int \frac{dx}{1+x^2}\] does that help?
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
yes. the Integration is the problem i'm having trouble with
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
okay first turn dy/ sec^2 y into cos^2 y dy you agree those are the same right?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
\[\implies \int \cos^2 y dy = \int \frac{dx}{1+x^2}\]
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
yes now i see.
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
wonderful
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
welcome
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
what about 1/(1+x^2)
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
trigonometric substitution
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
is it tan?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
arctan*
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
that's what I thought. Thanks
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
welcome
 one year ago
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