Question

How do you find the integral `int_0^(2)(10x)/sqrt(3-x^2)dx` ?

Answer 1

This is a classic case if what's called u-substitution. Meaning that you have to find a function ( u) and it's derivative (du) in the expression. Both the function and it's derivative may be hidden behind coefficients and confusing notation.

In this case, I might start by using exponents to rewrite the integral without fractions.

`int_0^2(10x)/sqrt(3-x^2)dx=int_0^2(3-x^2)^(-1/2)(10x)dx`

Now, I'm looking for a function and it's derivative. Here's where I notice that I have a second degree polynomial ( `3-x^2`) inside the exponent part, and I have a first degree polynomial (`10x`) elsewhere in the integral. In general a polynomial that starts with an x^2 will have a derivative that starts with an x. So:

If I decide that

`u=3-x^2`

then

`du=-2xdx`

but I don't have a `-2x` anywhere on the integral. I do have a `10x`.

I'm allowed to manipulate coefficients in an integral (manipulating variables is trickier, and sometimes not possible). So I need to manipulate the `10` in front of the x into being a `-2` in front of an x.

We start with:

`int_0^2(3-x^2)^(-1/2)(10x)dx`

We can pull a coefficient factor out of the integral entirely. We don't have to, but I find it makes for a more clear picture.

`10int_0^2(3-x^2)^(-1/2)(x)dx`

Now we would really like to put a `-2` in front of the x, so that we can match the `-2x` above. And we can put in in there. As long as we put it's reciprocal out in front to balance things out.

`10(1/(-2))int_0^2(3-x^2)^(-1/2)(-2x)dx`

=`-5int_0^2(3-x^2)^(-1/2)(-2x)dx`

Now we have our u and our du. But there's one other thing we need to be aware of.

This is a definite integral, and the 0 and the 2 represent x values, not u values. But we have a way to convert them,

`u=3-x^2`

So for `x=0`, we get `u=3-0^2`, or `u=3`.

For `x=2`, we get `u=3-2^2`, or `u=-1`.

Now we substitute all our x's for u's.

`-5int_0^2(3-x^2)^(-1/2)(-2x)dx=-5int_3^(-1)u^(-1/2)du`

`=-5(u^(1/2))/(1/2)|_3^(-1)=-10u^(1/2)|_3^(-1)`

`=-10sqrt(-1)-(-10sqrt(3))`

`=-10i+10sqrt(3)`, or

`10(sqrt(3)-i)`

Hope this helps.