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Math Help - Integral confusion

  1. #1
    Newbie Elvis's Avatar
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    Integral confusion

    I'm trying to find the integral \int\sec^2{\theta}\tan{\theta}\; {d\theta}. Here is what I did:

    Let u = \tan{\theta}. Then \dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C  \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C, which is right.

    My first question is, is there something wrong with the highlighted step? I have seen people raising questions about \dfrac{dy}{dx} being treated as if they were numbers.

    My second question is to do with when I tried to substitute u for \sec^2\theta and got different results than the above:

    Let u = \sec^2\theta. Then \dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C \; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C.

    Getting two different results for the same integral is something I can't grasp at all.
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  2. #2
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    trigonometric identities

    Quote Originally Posted by Elvis View Post
    I'm trying to find the integral \int\sec^2{\theta}\tan{\theta}\; {d\theta}. Here is what I did:

    Let u = \tan{\theta}. Then \dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C  \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C, which is right.

    My first question is, is there something wrong with the highlighted step? I have seen people raising questions about \dfrac{dy}{dx} being treated as if they were numbers.

    My second question is to do with when I tried to substitute u for \sec^2\theta and got different results than the above:

    Let u = \sec^2\theta. Then \dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C \; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C.

    Getting two different results for the same integral is something I can't grasp at all.
    sec^2(theta)-tan^2(theta)=1
    so when you substitute for any equation either of the values you will be left with a constant which goes with c.
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  3. #3
    Newbie Elvis's Avatar
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    Quote Originally Posted by Pulock2009 View Post
    sec^2(theta)-tan^2(theta)=1
    so when you substitute for any equation either of the values you will be left with a constant which goes with c.
    Could you please elaborate a bit further? I would lying if I say I understood what you have just said.
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  4. #4
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    skeeter's Avatar
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    Quote Originally Posted by Elvis View Post
    I'm trying to find the integral \int\sec^2{\theta}\tan{\theta}\; {d\theta}. Here is what I did:

    Let u = \tan{\theta}. Then \dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C  \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C, which is right.

    My first question is, is there something wrong with the highlighted step? I have seen people raising questions about \dfrac{dy}{dx} being treated as if they were numbers.

    My second question is to do with when I tried to substitute u for \sec^2\theta and got different results than the above:

    Let u = \sec^2\theta. Then \dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C \; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C.

    Getting two different results for the same integral is something I can't grasp at all.
    the two resulting antiderivatives differ only by a constant.
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Elvis View Post
    I'm trying to find the integral \int\sec^2{\theta}\tan{\theta}\; {d\theta}. Here is what I did:

    Let u = \tan{\theta}. Then \dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C  \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C, which is right.

    My first question is, is there something wrong with the highlighted step? I have seen people raising questions about \dfrac{dy}{dx} being treated as if they were numbers.

    My second question is to do with when I tried to substitute u for \sec^2\theta and got different results than the above:

    Let u = \sec^2\theta. Then \dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}} so that \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C \; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C.

    Getting two different results for the same integral is something I can't grasp at all.
    There is nothing wrong with the substitution you have made in the first operation.

    I hope you have learned trigonometry and you know that

    sec^{2}x - tan^{2}x =1

    so,

    your first answer is \frac{\tan^2\theta}{2}+C

    the second answer you have is \frac{\sec^2\theta}{2}+C

    = \frac{(1+\tan^2\theta)}{2}+C

    =  \frac{tan^{2}\theta}{2}+\frac{1}{2}+C

    Note that \frac{1}{2}+C= C' is also a constant. So it gives you \frac{tan^{2}\theta}{2}+C'

    So your answer by integration differ only by a constant
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  6. #6
    MHF Contributor harish21's Avatar
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    sorry. my answer got double posted
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