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Math Help - Some integrals

  1. #1
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    Some integrals

    I have to take a test that's a review of single-variable calculus, which I haven't really done in years. Most of the sample questions I've been able to do, but these are sticking me:

    \int^{\pi/2}_{0}\frac{\sin(x)}{1+\cos^{2}(x)}dx
    Now I know that this should be \tan^{-1}(\cos(x)) evaluated at the endpoints, but that's because I just looked that up. Is there a way to figure this out without simply having to know the derivative of \tan^{-1}?

    \int^{2}_{1}|x^{2}\sin(\pi x^{3})|dx
    Absolute values? Yikes. Do I just integrate this as if the absolutes weren't there, and then apply them when evaluating at the endpoints?

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Reisen View Post
    I have to take a test that's a review of single-variable calculus, which I haven't really done in years. Most of the sample questions I've been able to do, but these are sticking me:

    \int^{\pi/2}_{0}\frac{\sin(x)}{1+\cos^{2}(x)}dx
    Now I know that this should be \tan^{-1}(\cos(x)) evaluated at the endpoints, but that's because I just looked that up. Is there a way to figure this out without simply having to know the derivative of \tan^{-1}?
    do a substitution. u = cos(x)

    \int^{2}_{1}|x^{2}\sin(\pi x^{3})|dx
    Absolute values? Yikes. Do I just integrate this as if the absolutes weren't there, and then apply them when evaluating at the endpoints?
    no. the absolute values make a difference. you need to know how the graph without the absolute values behave. any interval that will give a negative integral will now give a positive one with the absolute values there. so you have to split the interval up. you have your work cut out for this one! see the graph below
    Attached Thumbnails Attached Thumbnails Some integrals-messy-absolute-value-integral.jpg  
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    do a substitution. u = cos(x)
    I meant in terms of figuring out that it's supposed to be \tan^{-1} without having to memorize the trig derivatives, but now that I think about it, that's kind of a silly question since I'm not doing to be deriving that stuff in the middle of the test.

    no. the absolute values make a difference. you need to know how the graph without the absolute values behave. any interval that will give a negative integral will now give a positive one with the absolute values there. so you have to split the interval up. you have your work cut out for this one! see the graph below
    Eep. Let's see, so the function equals zero whenever x^{3} is an integer. 1^{3}=1, 2^{3}=8, so I can split this up into
    -\int_{1}^{\sqrt[3]{2}}f(x)+\int_{\sqrt[3]{2}}^{\sqrt[3]{3}}f(x) - \int_{\sqrt[3]{3}}^{\sqrt[3]{4}}f(x) + \int_{\sqrt[3]{4}}^{\sqrt[3]{5}}f(x) - \int_{\sqrt[3]{5}}^{\sqrt[3]{6}}f(x) + \int_{\sqrt[3]{6}}^{\sqrt[3]{7}}f(x) - \int_{\sqrt[3]{7}}^{2}f(x)

    and evaluate that. The antiderivative looks like it should be -\frac{1}{3\pi}\cos(\pi x^{3}), so this won't be super difficult, just tedious. Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Reisen View Post
    I meant in terms of figuring out that it's supposed to be \tan^{-1} without having to memorize the trig derivatives, but now that I think about it, that's kind of a silly question since I'm not doing to be deriving that stuff in the middle of the test.
    nope, in this case, memorizing the trig derivative is the way to go. but i do not think your question is silly in general. it is good to know how to derive things. in general, it makes life easier for you, helps to cut down on how much memory you have to use. and if you're fast enough, it doesn't slow you down much in tests. at least, not as far as elementary calculus is concerned... usually

    Eep. Let's see, so the function equals zero whenever x^{3} is an integer. 1^{3}=1, 2^{3}=8, so I can split this up into
    -\int_{1}^{\sqrt[3]{2}}f(x)+\int_{\sqrt[3]{2}}^{\sqrt[3]{3}}f(x) - \int_{\sqrt[3]{3}}^{\sqrt[3]{4}}f(x) + \int_{\sqrt[3]{4}}^{\sqrt[3]{5}}f(x) - \int_{\sqrt[3]{5}}^{\sqrt[3]{6}}f(x) + \int_{\sqrt[3]{6}}^{\sqrt[3]{7}}f(x) - \int_{\sqrt[3]{7}}^{2}f(x)

    and evaluate that. The antiderivative looks like it should be -\frac{1}{3\pi}\cos(\pi x^{3}), so this won't be super difficult, just tedious. Thanks!
    yes
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