# definite integration no.8

• November 7th 2007, 12:36 AM
afeasfaerw23231233
definite integration no.8
p218 ex22a q 31
• November 7th 2007, 03:26 AM
topsquark
Quote:

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 $1 - sin(x)$.

-Dan
• November 7th 2007, 03:59 AM
Krizalid
$\int {\frac{{1 - t^2 }}{{1 + t^2 }}\,dt} = \int {\frac{{2 - (1 + t^2 )}}{{1 + t^2 }}\,dt} = 2\int {\frac{1}{{1 + t^2 }}\,dt} - \int {dt} .$

I think this way takes less time.