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Math Help - Two trig integrals

  1. #1
    Junior Member symstar's Avatar
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    Two trig integrals

    First:
    \int \sin^3x \cos^3x dx

    I used u=\cos x \text{    }du=-\sin x dx

    My answer was : =\tfrac{1}{6}cos^{6} x-\tfrac{1}{4}\cos^{4} x + C

    Apparently, the answer given was with u=\sin x which gives the answer: =\tfrac{1}{4}\sin^4x-\tfrac{1}{6}\sin^6x + C

    So, are these answers the same or was I just wrong?


    Second:
    \int^1_{-1} \frac{(x+2)^2}{x^2+1}dx
    x=\tan\theta \text{   }dx=\sec^2\theta d\theta
    =\int^1_{-1} \tan^2\theta d\theta + 4\int^1_{-1} \tan\theta d\theta + 4\int^1_{-1} d\theta
    =\int^1_{-1}\frac{\sin^2\theta}{\cos^2\theta} d\theta + 4*\ln{\left|\sec\theta\right| \Big|^1_{-1}} + 4*\theta \Big|^1_{-1}
    =\int^1_{-1}\frac{1-cos^2\theta}{cos^2\theta}d\theta + 4*\ln{(\sqrt{x^2+1})\Big|^1_{-1}} + 4*\tan^{-1}x\Big|^1_{-1}
    =\int^1_{-1}\sec^2\theta d\theta - \int^1_{-1} d\theta + 4(\ln{(\sqrt{2})}-\ln{(\sqrt{2})}) + 4(\tan^{-1}(1)-\tan^{-1}(-1))
    =\tan\theta\Big|^1_{-1} - \theta \Big|^1_{-1} + 4(\tfrac{\pi}{4}-\tfrac{3\pi}{4})
    =2-(\tfrac{\pi}{4}-\tfrac{3\pi}{4}) + 4(-\tfrac{\pi}{2})
    =2-\tfrac{3\pi}{2}

    The answer given is: =2+\tfrac{3\pi}{2}

    Where did the sign change come from or is the given answer wrong?
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  2. #2
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    for the 1st problem, you're ok ... the two expressions only differ by a constant.



    \int_{-1}^1 \frac{x^2 + 2x + 4}{x^2 + 1} dx =

    \int_{-1}^1 \frac{x^2}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2 + 1} dx =<br />

    \int_{-1}^1 1 - \frac{1}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2+1} \, dx =

    \int_{-1}^1 1 + \frac{2x}{x^2+1} + \frac{3}{x^2+1} \, dx =

    [x]_{-1}^1 + [\ln(x^2+1)]_{-1}^1 + 3[\arctan{x}]_{-1}^1 =

    2 + \frac{3\pi}{2}
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  3. #3
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    The first one is correct. As for the second one, I can't really check your work but perhaps you should re-evaluate it. Alternatively, if you expand the numerator and use polynomial division, you will get a much simpler integral to evaluate (and the answer would be the one given in the book).
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  4. #4
    Junior Member symstar's Avatar
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    Yeah, but where's the sport in doing it the easy way .
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  5. #5
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    "sport" only counts when chasing members of the opposite sex.

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