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Math Help - 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) \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) \int tan^2(x)

    P.S
    \int tan^2(x)dx = \int (sec^2(x) - 1)dx
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    1) \int tan^2(x)
    \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...

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

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

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

    Kind regards

    \chi \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 \int\sec^2{x}dx ?

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

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

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

    Kind regards

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

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

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

    dv = sin(x) v = -cos(x)

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

    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:

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

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

    dv = sin(x) v = -cos(x)

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

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

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

    u = \sin{x}

    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

    u = \sin{x}

    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:

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


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

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