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

Oct 1, 2014

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.

${\int}_{0}^{2} \frac{10 x}{\sqrt{3 - {x}^{2}}} \mathrm{dx} = {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(10 x\right) \mathrm{dx}$

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 ($10 x$) 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

$\mathrm{du} = - 2 x \mathrm{dx}$

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

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.

${\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(10 x\right) \mathrm{dx}$

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.

$10 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(x\right) \mathrm{dx}$

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

$10 \left(\frac{1}{- 2}\right) {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx}$

=$- 5 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx}$

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.

$- 5 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx} = - 5 {\int}_{3}^{- 1} {u}^{- \frac{1}{2}} \mathrm{du}$

$= - 5 \frac{{u}^{\frac{1}{2}}}{\frac{1}{2}} {|}_{3}^{- 1} = - 10 {u}^{\frac{1}{2}} {|}_{3}^{- 1}$

$= - 10 \sqrt{- 1} - \left(- 10 \sqrt{3}\right)$

$= - 10 i + 10 \sqrt{3}$, or

$10 \left(\sqrt{3} - i\right)$

Hope this helps.