please can someone help me intergrate these please :
Tan -1x
1 + x(squared) dx
and.
4cos3s ds
many thanks people
i suppose the first question is this..
$\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx$
set $\displaystyle u = \tan ^{-1} x \Rightarrow du = \frac{1}{1+x^2} \, dx$ and that should be fine..
for the second one, set $\displaystyle u = 3s \Rightarrow du = 3ds$ and you can do it all the way..
ok..
$\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx$
set $\displaystyle u = \tan ^{-1} x \Rightarrow du = \frac{1}{1+x^2} \, dx$
then $\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\tan^{-1} x)^2}{2} + C$..
is that ok?