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Thread: Trgnometric integrals

  1. #1
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    Trgnometric integrals

    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
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    Quote Originally Posted by radioheadfan View Post
    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
    Use the chain rule.

    $\displaystyle y = \tan^4{x} = (\tan{x})^4$.


    Let $\displaystyle u = \tan{x}$ so that $\displaystyle y = u^4$.


    $\displaystyle \frac{du}{dx} = \sec^2{x}$


    $\displaystyle \frac{dy}{du} = 4u^3$

    $\displaystyle = 4\tan^3{x}$.


    Therefore

    $\displaystyle \frac{dy}{dx} = 4\sec^2{x}\tan^3{x}$.
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  3. #3
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    Quote Originally Posted by radioheadfan View Post
    How would you evaluate this indefinite integral?
    tan^4x dx

    I began with Seperating it into two tans. Both to the 2nd power.
    Then I got lost. How do I get to my answer?
    $\displaystyle \int \tan^4{x} \, dx$

    $\displaystyle \int \tan^2{x}(\sec^2{x} - 1) \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx$

    first integral ... $\displaystyle u =tan{x}$ ... $\displaystyle du = \sec^2{x} \, dx$

    $\displaystyle \frac{\tan^3{x}}{3} - \tan{x} + x + C$
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    Quote Originally Posted by skeeter View Post
    $\displaystyle \int \tan^4{x} \, dx$

    $\displaystyle \int \tan^2{x}(\sec^2{x} - 1) \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx$

    $\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx$

    first integral ... $\displaystyle u =tan{x}$ ... $\displaystyle du = \sec^2{x} \, dx$

    $\displaystyle \frac{\tan^3{x}}{3} - \tan{x} + x + C$

    Oops, I read that it had to be differentiated. LOL.
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