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Math Help - Trigonometric Substitution Calc. II Help Please!

  1. #1
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    Trigonometric Substitution Calc. II Help Please!

    integral 8/((4x^2+1)^2) dx

    I know you have to let x=1/2tanu
    therefore dx = 1/2 sec^2 u du

    But I get a wrong answer! Somewhere along the way I get messed up.

    Also:

    integral e^t/(sqrt(e^(2t)+9)) dt from 0 to ln4

    Use a substitution like u=e^t to simplify the problem, then let u = 3tant
    and so du = 3sec^2t

    I also get a wrong answer for this one. Be careful with the limits.
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  2. #2
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    Quote Originally Posted by joepinsley View Post
    integral 8/((4x^2+1)^2) dx

    I know you have to let x=1/2tanu
    therefore dx = 1/2 sec^2 u du
    \int \frac{4\sec^2{u}}{(\tan^2{u}+1)^2} \, du

    \int \frac{4\sec^2{u}}{\sec^4{u}} \, du

    \int 4\cos^2{u} \, du

    2\int 1 + \cos(2u) \, du

    2u + \sin(2u) + C

    u = \arctan(2x) ...

    2\arctan(2x) + \sin[2 \arctan(2x)] + C

    2\arctan(2x) + 2\sin[\arctan(2x)]\cos[\arctan(2x)] + C

    2\arctan(2x) + 2 \cdot \frac{2x}{\sqrt{4x^2+1}} \cdot \frac{1}{\sqrt{4x^2+1}} + C

    2\arctan(2x) +  \frac{4x}{4x^2+1}  + C
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