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Thread: Simple Integration problem

  1. #1
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    Simple Integration problem

    Hi

    Can someone tell me how i would integrate the following question:

    1) $\displaystyle \int tan^2(x)$

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi

    Can someone tell me how i would integrate the following question:

    1) $\displaystyle \int tan^2(x)$

    P.S
    $\displaystyle \int tan^2(x)dx$ = $\displaystyle \int (sec^2(x) - 1)dx$
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    1) $\displaystyle \int tan^2(x)$
    $\displaystyle \int{\tan^2(x)}\;{dx} = \int{\left\{\sec^2(x)-1\right\}}\;{dx}$

    EDIT: Bit late.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Because is...

    $\displaystyle \tan^{2} x = \sin x \cdot \frac{\sin x}{\cos^{2} x}$ (1)

    ... You can integrate by parts and obtain...

    $\displaystyle \int \tan^{2} x \cdot dx = \frac{\sin x}{\cos x} - \int dx = \tan x - x + c$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Can you show me how you have done it, because i still cannot get the right answer.
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    Can you show me how you have done it, because i still cannot get the right answer.
    Show your calculation.

    Do you know $\displaystyle \int\sec^2{x}dx$ ?

    and $\displaystyle \int{dx}$ ?
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  7. #7
    MHF Contributor chisigma's Avatar
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    Integration by parts:

    $\displaystyle \int f(x)\cdot g(x)\cdot dx = f(x)\cdot G(x) - \int G(x)\cdot f^{'} (x)\cdot dx$ (1)

    ... where $\displaystyle G(*)$ is a primitive of $\displaystyle g(*)$. In our case is $\displaystyle f(x)= \sin x$ and $\displaystyle g(x)= \frac{\sin x}{\cos^{2} x}$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  8. #8
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    Ok this is what i have done:

    $\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$

    $\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$

    $\displaystyle dv = sin(x) v = -cos(x)$

    $\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$

    $\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$

    Someone tell me where my mistake is.
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  9. #9
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Ok this is what i have done:

    $\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$

    $\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$

    $\displaystyle dv = sin(x) v = -cos(x)$

    $\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$

    $\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$

    Someone tell me where my mistake is.
    you picked the wrong u and v

    $\displaystyle u = \sin{x}$

    $\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$
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  10. #10
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    Quote Originally Posted by 11rdc11 View Post
    you picked the wrong u and v

    $\displaystyle u = \sin{x}$

    $\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$
    how can i find what v is equal to?
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  11. #11
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    By doing Integration by Substitution:

    $\displaystyle \int \frac{\sin{x}}{\cos^2{x}}dx $


    $\displaystyle u = \cos x \Longrightarrow du = -\sin x dx $

    $\displaystyle = -\int \frac{1}{u^2}du $
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