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Math Help - definite integration no.8

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    definite integration no.8

    p218 ex22a q 31
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p218 ex22a q 31
    You can put this into a more "standard" form by multiplying the top and bottom of the integrand by 1 - sin(x).

    -Dan
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    \int {\frac{{1 - t^2 }}{{1 + t^2 }}\,dt} = \int {\frac{{2 - (1 + t^2 )}}{{1 + t^2 }}\,dt} = 2\int {\frac{1}{{1 + t^2 }}\,dt} - \int {dt} .

    I think this way takes less time.
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