Question

How do you find the integral `int t^2(t^3+4)^(-1/2)dt` using substitution?

Answer 1

`2/3(t^3+4)^(1/2)+C` or `2/3sqrt(t^3+4)+C`

Explanation:

The key to solving any integral is to see what "type" of integral it could classify as. When I see an integral I try to ask myself if it is a substitution, integration by parts, trigonometric, trig sub, or partial fractions integral. In order to know whether or not to use the substitution method is whether or not the integral has both a function and its derivative present in the integral.

For this integral: `intt^2(t^3+4)^(-1/2)dt`, there is a `t^3` and a `t^2` which hints that we may want to incorporate `t^3` into our substitution since its derivative is present.

Let's do just that.
`intt^2/(t^3+4)^(1/2)dt`

Let `u=t^3+4`
so, `du=3t^2dt` or `1/3du=t^2dt`

`1/3int1/u^(1/2)du=1/3intu^(-1/2)du=(1/3)(2u^(1/2))+C=2/3u^(1/2)+C`

Putting `t` back in, we get:
`2/3(t^3+4)^(1/2)+C`

or `2/3sqrt(t^3+4)+C`

Answer 2

`intt^2(t^3+4)^(-1/2)dt=2/3(t^3+4)^(1/2)+C`

Explanation:

Because the function outside the bracket is a constant`xx`the derivative of the bracket , it suggests that we might be able to do this by 'inspection'.

`intt^2(t^3+4)^(-1/2)dt`

using the power rule for integration let us 'guess' that teh integral is of teh form

`y=(t^3+4)^(1/2)`

differentiate this using the chain rule:

`(dy)/(dx)=1/2xx3t^2(t^3+4)^(-1/2)`

comparing this with the original question we see that the only difference is the constant `3/2` so we adjust our 'guess' by a suitable value.

`:.intt^2(t^3+4)^(-1/2)dt=2/3(t^3+4)^(1/2)+C`

If one can spot integrals by inspection it can save a lot of work!.