p218 ex22a q 31
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Originally Posted by afeasfaerw23231233 p218 ex22a q 31 You can put this into a more "standard" form by multiplying the top and bottom of the integrand by $\displaystyle 1 - sin(x)$. -Dan
$\displaystyle \int {\frac{{1 - t^2 }}{{1 + t^2 }}\,dt} = \int {\frac{{2 - (1 + t^2 )}}{{1 + t^2 }}\,dt} = 2\int {\frac{1}{{1 + t^2 }}\,dt} - \int {dt} .$ I think this way takes less time.
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