Question

How do I integrate the following equation:
x(r^2-x^2)^(1/2)?
r=.5

Answer 1

\[\int\limits_{0}^{.5}x \sqrt{r^{2}-x^{2}}\]
Step by step please.

Answer 2

What is the variable of integration?

Answer 3

it's x

Answer 4

there is a very obvious substitution here

Answer 5

notice that \(\dfrac{d}{dx}\left(r^2-x^2\right) = -2x\)

Answer 6

but isn't there a way like trigonometric sustitution to solve this problem? Because the final result gives me 0 after giving it values.

Answer 7

There's a way to do it with "u-substitution" or "trig-substitution". You can do whatever you like here.

Answer 8

As rational said, it is a substitution integral, pretty easy to spot by naked eye, because square roots are very hard to integrate by their own so we will choose what's inside of it to make more friendly to integrate.
So if we have:
\[\int\limits_{0}^{5}x \sqrt{r^2 - x^2 }dx\]
Don't forget the diferential at the end.
So, let's name the new variable "U" and let's take what's inside the integral so we can make it a simple sqrt x one.
\[u= r^2 - x^2\]
And derivating:
\[\frac{ du }{ dx }= -2x\]
So far we have:
\[\int\limits_{}^{}x \sqrt{u}dx\]
This is not integrable just yet, because we have an "x" and a "dx" there, we need to make it all in terms of "u" and "du", so let's first take the change of variable and solve for x:
\[u= r^2 - x^2 \]
\[x=\sqrt{r^2-u}\]
And, the "du":
\[\frac{ du }{ dx }=-2x\]
To make a "du" appear, we will solve it for "dx":
\[dx=\frac{ du }{ -2x }\]
So far, we have:
\[\int\limits_{}^{}(\sqrt{r^2-u})\sqrt{u} \frac{ du }{ -2(\sqrt{r^2 - u}) }\]
We can simplify those annoying Sqrt(r^2 - u) so we get:
\[\int\limits_{}^{}\sqrt{u}\frac{ du }{ -2 }\]
or:
\[\int\limits_{}^{}\frac{ \sqrt{u}du }{ -2 }\]
Now, we have defined a whole new function, so we have to take the change of variable again, and :
\[r^2 - (0)^2 = r^2\]
\[r^2-(5)^2 = r^2 - 25\]
So, the new integral will be:
\[\int\limits_{r^2}^{r^2 - 25} \frac{ \sqrt{u}du }{ -2 }\]

Answer 9

If you're hell bent on doing trig sub, I'll just go ahead and show you how you might figure out which substitution to make for x: \[\large \sin^2 \theta + \cos^2 \theta = 1 \\ \large r^2 \sin^2 \theta + r^2 \cos^2 \theta = r^2 \\ \large r^2 \cos^2 \theta = r^2 -r^2 \sin^2 \theta \\ \therefore x=r \sin \theta \]