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    Please solve this for me ...

    [Intergral sign] (x^2)(sec^2)(tanx)dx

    I've put the brackets in so its easier to read; if you don't want them then it is: x^2sec^2tanx dx

    And I'll put it in words too: Intergral of x squared, sec squared x, tanx with respect to x.

    It's intergration by parts I believe
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    Do you know an identity relating sec and tan that might tidy it up a bit?

    Spoiler:
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    sec^{2}(x) = 1 + tan^2(x)


    (Note: I haven't actually tried it, that was the first thing I thought when I saw it)
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    (Original post by JamesyB)
    Do you know an identity relating sec and tan that might tidy it up a bit?

    Spoiler:
    Show
    sec^{2}(x) = 1 + tan^2(x)


    (Note: I haven't actually tried it, that was the first thing I thought when I saw it)
    Hmm...don't see where you can go from there to be honest :confused:
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    Notice that \sec^2 x \tan x is similar to the derivative of \sec^2 x, which hints that you might want to use integration by parts. You'll need to do it twice; once with u=x^2, \frac{dv}{dx}=\sec^2 x \tan x, and then again with u=x and \frac{dv}{dx}=\text{other stuff}.
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    (Original post by nuodai)
    Notice that \sec^2 x \tan x is similar to the derivative of \sec^3 x, which hints that you might want to use integration by parts. You'll need to do it twice; once with u=x^2, \frac{dv}{dx}=\sec^2 x \tan x, and then again with u=x and \frac{dv}{dx}=\text{other stuff}.
    I don't know what the deritive of sec^3 x is? And I don't know how to integrate sec^2xtanx, unless I have to do it by parts which I suppose may work.
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    (Original post by A level Az)
    I don't know what the deritive of sec^3 x is? And I don't know how to integrate sec^2xtanx, unless I have to do it by parts which I suppose may work.
    Woops sorry, I meant the derivative of \sec^2 x. If don't know what it is, use the chain rule: if u=\sec x and y=u^2 =\sec^2 x then what is \frac{dy}{dx}?

    [But on another note, if you can do integration by parts, you can probably differentiate \sec^2 x and \sec^3 x.]
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    (Original post by nuodai)
    Woops sorry, I meant the derivative of \sec^2 x. If don't know what it is, use the chain rule: if u=\sec x and y=u^2 =\sec^2 x then what is \frac{dy}{dx}?

    [But on another note, if you can do integration by parts, you can probably differentiate \sec^2 x and \sec^3 x.]
    Ok thanks I'll try to get it to come out. And by the way, what would you say is the intergral of (x-1)? Is it right to say it is 0.5x^2 - x, or does it have to be (x-1)^2 all over 2? There is this other question and its mark scheme is very unclear.
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    Re \int (x-1)\,dx Either is fine. They just differ by a constant.
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    (Original post by DFranklin)
    Re \int (x-1)\,dx Either is fine. They just differ by a constant.
    Well, when its part of a question, like for example, the intergral of (x-1)lnx, with the boundaries 3 and 1, then it really messes things up. If you intergrate to 0.5x^2 - x then you miss the constant (1) so the answer is wrong. The correct way to intergrate for that question is (x-1)^2 all over 2, but I didn't realise that . Do you have to take it as a whole bracket when there is something else being multiplied with the bracket?
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    (Original post by A level Az)
    Well, when its part of a question, like for example, the intergral of (x-1)lnx, with the boundaries 3 and 1, then it really messes things up.
    No it doesn't. When you integrate between definite boundaries, you're effectively adding the constant and subtracting it again, so it doesn't matter what it is. You get the same answer either way.

    If you're getting different answers, then you're doing something wrong. I suggest you post your working.
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    (Original post by DFranklin)
    No it doesn't. When you integrate between definite boundaries, you're effectively adding the constant and subtracting it again, so it doesn't matter what it is. You get the same answer either way.

    If you're getting different answers, then you're doing something wrong. I suggest you post your working.
    Worked it out and you're right, thanks for the help
 
 
 
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