- Thread starter
- #1
karush
Well-known member
- Jan 31, 2012
- 2,861
Compute
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1-4t^2}dt$
Thot I could solve by using the log rule for integration namely:
$\displaystyle\int\frac{u'}{u} = ln|u|+C$
for such $u=1-4t^2$ and $u'=-8t$
but then $\displaystyle\frac{1}{2}\int_0^\frac{1}{2}\frac{-8t}{1-4t^2}dt$ isn't going to work due to the t in the numerator....so what shud I do....
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1-4t^2}dt$
Thot I could solve by using the log rule for integration namely:
$\displaystyle\int\frac{u'}{u} = ln|u|+C$
for such $u=1-4t^2$ and $u'=-8t$
but then $\displaystyle\frac{1}{2}\int_0^\frac{1}{2}\frac{-8t}{1-4t^2}dt$ isn't going to work due to the t in the numerator....so what shud I do....
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