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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?
 2 years ago

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

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

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

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

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

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

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

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

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

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