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Math Help - Integral

  1. #1
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    Integral

    Don't know how to work out the following integral.
    Substituting cos x^5 = t does not seem to work.
    Attached Thumbnails Attached Thumbnails Integral-integraal.jpg  
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  2. #2
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    This looks much harder than it is.

    Let u=x^5, du=5x^4dx

    Another way of solving for "du" is \frac{1}{5}du=x^4dx. But we don't want just x^4dx, we want 2x^4, so multiply by 2 on both sides to get \frac{2}{5}du=2x^4. Now we're getting somewhere.

    So plug everything back in. Our new integral in terms of u is...

    \frac{2}{5} \int \frac{1}{\cos^2(u)}du

    Now what's \frac{1}{\cos^2(x)}?? \sec^2(x).

    See it now?
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    Quote Originally Posted by totalnewbie View Post
    Don't know how to work out the following integral.
    Substituting cos x^5 = t does not seem to work.
    Let,
    u=x^5 then u'=5x^4
    Thus,
    \frac{2}{5} \int \frac{5x^4}{\cos^2 x^5}dx
    Is,
    \frac{2}{5} \int \sec^2u u' dx=\frac{2}{5}\tan u+C=\frac{2}{5}\tan (x^5)+C
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    What is SEC ? Don't find that kind of word from the dictionary.
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    Should I substitute again ?
    I am not able to get rid of x from denominator
    Attached Thumbnails Attached Thumbnails Integral-integraal2.jpg  
    Last edited by totalnewbie; December 3rd 2006 at 09:21 AM. Reason: previous graph was inaccurate
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    Quote Originally Posted by totalnewbie View Post
    What is SEC ? Don't find that kind of word from the dictionary.
    Secant
    ----
    Your other problem that you posted cannot be done.
    It leads to,
    \int e^{x^2}dx which cannot be done.
    Last edited by ThePerfectHacker; December 3rd 2006 at 07:21 AM.
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  7. #7
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    Quote Originally Posted by totalnewbie View Post
    Should I substitute again ?
    I am not able to get rid of x from denominator
    Hello,

    your problem looks harder than it is actually:

    \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{\frac{1}{x} \cdot (\ln(x)+2)^{-\frac{2}{5}}dx}

    let u=\ln(x)+2. Then
    \frac{du}{dx}=\frac{1}{x}. Thus du=\frac{1}{x} dx.
    Now plug in these terms into the original integral:

    \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{u^{-\frac{2}{5}} du}


    I'll leave the rest for you.

    (For confirmation only: \frac{5}{3}(\ln(x)+2)^{\frac{3}{5}}

    EB
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    Quote Originally Posted by earboth View Post
    Hello,

    your problem looks harder than it is actually:

    \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{\frac{1}{x} \cdot (\ln(x)+2)^{-\frac{2}{5}}dx}

    let u=\ln(x)+2. Then
    \frac{du}{dx}=\frac{1}{x}. Thus du=\frac{1}{x} dx.
    Now plug in these terms into the original integral:

    \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{u^{-\frac{2}{5}} du}


    I'll leave the rest for you.

    (For confirmation only: \frac{5}{3}(\ln(x)+2)^{\frac{3}{5}}

    EB
    The problem is that he has,
    \frac{1}{x^5} not \frac{1}{x}
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    The problem is that he has,
    \frac{1}{x^5} not \frac{1}{x}
    Hi TPH,

    the image of the problem i can see shows a multiplication sign between the x and the 5th root. So I believe that I read it correctly.

    EB
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    \int \arctan{x}dx

    Any ideas ?
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  11. #11
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    Quote Originally Posted by totalnewbie View Post
    \int \arctan{x}dx

    Any ideas ?
    One thing I hate about integral is checking differenciability intervals and domains. Since I am lazy here was my approach.

    \int \tan^{-1} x dx
    Express as,
    \int (1+x^2) \cdot \frac{\tan^{-1} x}{1+x^2} dx
    Now,
    1+x^2=\sec ^2 (\tan^{-1} x)
    Thus,
    \int \sec^2 (\tan^{-1} x) \cdot \tan^{-1} x\cdot \frac{1}{1+x^2} dx
    Let,
    f(x)=x\sec^2x
    g(x)=\tan^{-1} x
    g'(x)=1/(1+x^2)
    Thus,
    \int (f \circ g)\cdot g' dx=F\circ g+C
    Where,
    F is an anti-derivative of x\sec^2 x
    So the problem reduces to finding,
    \int x\sec^2 xdx
    Which should be doable by integration by parts.
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  12. #12
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    \int \frac{xdx}{1+x^2}=\frac{1}{2}\int\frac{d(1+x^2)}{1  +x^2}=\frac{1}{2}ln|1+x^2|+C

    Where does \frac{1}{2} come from ?
    Where does d(1+x^2) come from ?
    If \int\frac{dx}{1+x^2}=arctanx+C how is it possible that my answer has ln ?
    Last edited by totalnewbie; December 7th 2006 at 06:03 AM.
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  13. #13
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    Quote Originally Posted by totalnewbie View Post
    \int \frac{xdx}{1+x^2}=\frac{1}{2}\int\frac{d(1+x^2)}{1  +x^2}=\frac{1}{2}ln|1+x^2|+C

    Where does \frac{1}{2} come from ?
    Where does d(1+x^2) come from ?
    If \int\frac{dx}{1+x^2}=arctanx+C how is it possible that my answer has ln ?
    Becuse you are substituting x^2+1.
    Its derivative is 2x. Note the 2. In the numerator you just have x. So to get the 2. You multiply and divide by 2 (that is 1/2). That is where it comes from.
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  14. #14
    TD!
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    I would start off with integration by parts:

    <br />
\int {\arctan x} dx = x\arctan x - \int {\frac{x}{{1 + x^2 }}dx} <br />

    Now, the second integral is easy:

    <br />
\int {\frac{x}{{1 + x^2 }}dx}  = \frac{1}{2}\int {\frac{1}{{1 + x^2 }}d\left( {1 + x^2 } \right) = \frac{1}{2}\ln \left| {1 + x^2 } \right|}  + C<br />

    So combined:

    <br />
\int {\arctan x} dx = x\arctan x - \frac{1}{2}\ln \left| {1 + x^2 } \right| + C<br />
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  15. #15
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    Quote Originally Posted by totalnewbie View Post
    \int \arctan{x}dx

    Any ideas ?
    try \tan(u)=x

    RonL
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