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    ∫(sin²2x)(tan2x)dx, u = cos2x

    1.differentiated u then replaced dx with du/-sin2x
    2.changed sin²2x to 1 - cos²2x
    3. changed tan2x with sin2x/cos2x

    simplified and i got (cos2x)²/2 - in(cos2x) +c but the answer in the book is (cos2x)²/4 - in(cos2x)/2 +c

    i think my mistake is got to do with the 2x in the functions
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    (Original post by man111111)
    ∫(sin²2x)(tan2x)dx, u = cos2x

    1.differentiated u then replaced dx with du/-sin2x
    but dx = \dfrac{du}{-2\sin(2x)}...
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    (Original post by RDKGames)
    but dx = \dfrac{du}{-2\sin(2x)}...
    thanks im stuck on this question ∫ 2/e^2x+4 dx , u = e^2x+4

    1.differentiated u then replaced dx with du/2e^2x
    2.changed e^2x+4 to u
    3. simplified to get ∫1/(u)(u-4)du then i said this is equal to in((u)(u-4)) but this is no where near the answer in the book
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    (Original post by man111111)
    thanks im stuck on this question ∫ 2/e^2x+4 dx , u = e^2x+4

    1.differentiated u then replaced dx with du/2e^2x
    2.changed e^2x+4 to u
    3. simplified to get ∫1/(u)(u-4)du then i said this is equal to in((u)(u-4)) but this is no where near the answer in the book
    What's your reason behind saying that \displaystyle \int \dfrac{1}{u(u-4)}.du = \ln (u(u-4))? This is important to get out the way before you make this common silly mistake more and more.
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    (Original post by RDKGames)
    What's your reason behind saying that \displaystyle \int \dfrac{1}{u(u-4)}.du = \ln (u(u-4))? This is important to get out the way before you make this common silly mistake more and more.
    i forgot to say that after this i replaced the u with e^2x+4
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    (Original post by man111111)
    i forgot to say that after this i replaced the u with e^2x+4
    Ok... but that doesn't answer my question to you..
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    (Original post by RDKGames)
    Ok... but that doesn't answer my question to you..
    so what i have understood from my teacher is that when 1 is the numerator then you can integrate it to In(denominator)

    this is the reason for why i have said that
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    (Original post by man111111)
    so what i have understood from my teacher is that when 1 is the numerator then you can integrate it to In(denominator)
    That's not correct, might want to follow that up with your teacher.
    By that reasoning, we get that \displaystyle \int \frac{1}{x^2}.dx = \ln(x^2) +c but we also get that \displaystyle \int x^{-2}.dx = -x^{-1} + c so the two RHS's are equal, when they're actually not.

    The only time you integrate into log is by the following proposition: \displaystyle \int \dfrac{f'(x)}{f(x)} .dx = \ln |f(x)| +c.

    Clearly, the requirement here is that the numerator is the derivative of the denominator. In your case, you have \dfrac{1}{u(u-4)} = \dfrac{1}{u^2-4u} but clearly the numerator is not the derivative of the denominator.

    Instead, you need to split \dfrac{1}{u(u-4)} into partial fractions, and then the book's answer is apparent.
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    (Original post by RDKGames)
    That's not correct, might want to follow that up with your teacher.
    By that reasoning, we get that \displaystyle \int \frac{1}{x^2}.dx = \ln(x^2) +c but we also get that \displaystyle \int x^{-2}.dx = -x^{-1} + c so the two RHS's are equal, when they're actually not.

    The only time you integrate into log is by the following proposition: \displaystyle \int \dfrac{f'(x)}{f(x)} .dx = \ln |f(x)| +c.

    Clearly, the requirement here is that the numerator is the derivative of the denominator. In your case, you have \dfrac{1}{u(u-2)} = \dfrac{1}{u^2-2u} but clearly the numerator is not the derivative of the denominator.

    Instead, you need to split \dfrac{1}{u(u-2)} into partial fractions, and then the book's answer is apparent.
    thanks for your detailed reply. this question is in the C3 section so is there another way of answering this question rather than using partial fractions?
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    (Original post by man111111)
    thanks for your detailed reply. this question is in the C3 section so is there another way of answering this question rather than using partial fractions?
    Without it, you can also expand on the denominator and complete the square on it, then use the integral in your formula booklet for \displaystyle \int \frac{1}{x^2-a^2} .dx = \frac{1}{2a} \ln |\frac{x-a}{x+a}|+c
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    (Original post by RDKGames)
    Without it, you can also expand on the denominator and complete the square on it, then use the integral in your formula booklet for \displaystyle \int \frac{1}{x^2-a^2} .dx = \frac{1}{2a} \ln |\frac{x-a}{x+a}|+c
    thank you
 
 
 
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