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  1. B

    Integration using Trig Substitution

    My mistake, you're right. Thanks!
  2. B

    Integration using Trig Substitution

    Yes, I'm fairly certain that is a standard integral, and it comes out to: sin^{-1}y + C But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to...
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    Integration using Trig Substitution

    So I should end up with: \int sec\Theta d\Theta} Correct? then I would get ln \left| sec\Theta+tan\Theta \right| + C
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    Integration using Trig Substitution

    as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ? I can get to: \int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}} But after that it makes no sense, I just get back to the original equation when I sub. again.
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    Integration using Trig Substitution

    Homework Statement \int \frac{cosx dx}{\sqrt{1 + sin^{2}x}} Homework Equations Expression: \sqrt{a^{2} + x^{2}} Substitution: x = a*tan\Theta Identity: 1 + tan^{2}\Theta = sec^{2}\Theta The Attempt at a Solution I have tried using Trig Substitution, but I end up getting an equation much...
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    Trifunction Derivatives problem

    The quotient Rule is: y= u/v y'= [(vu'-uv')/(v^2)] The product Rule is: y= uv y'= (vu'+uv') the chain Rule is: y=u^n y'=(n)(u^n-1)(u')
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