How do you integrate x3+x2dx using trigonometric substitution?

1 Answer
Jan 24, 2017

x3+x2dx=13(3+x2)3+x2+C

Explanation:

You do not really need a trigonometric substitution here. As:

d(3+x2)=2xdx

you can just substitute:

t=3+x2

and have:

x3+x2dx=12tdt=13t32+C=13(3+x2)3+x2+C

However if you want to perform a trigonometric substitution, the right substitution when under the square root there is a sum of square is:

x=3tant with t(π2,π2)

dx=3dtcos2t

In this case we have:

x3+x2dx=3tant3+3tan2tdtcos2t=33tantcos2t1+tan2tdt

we now use the trigonometric identity:

1+tan2t=1cos2t

and have:

x3+x2dx=33tantcos2t1cos2tdt

Then we note that cost>0 in the interval t(π2,π2), so:

x3+x2dx=33tantcos3tdt=33sintcos4tdt=33dcostcos4t=3cos3t+C

To substitute back x, we need to express cost in terms of tant:

cost=11+tan2t=11+x23=33+x2

so that:

x3+x2dx=3(33+x2)3+C=13(3+x2)3+x2+C