1. ## integration power rule

the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?

2. Originally Posted by dataspot
the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
Consider the derivative of the inside of the radical?

Or more conservatively

Let $\displaystyle \psi=4-x^2$
so then $\displaystyle d\psi=-2xdx$

now looking at our integral we have

$\displaystyle \frac{-1}{2}\int\frac{d\psi}{\sqrt{\psi}}=\frac{-1}{2}\cdot{2\sqrt{\psi}}$

Now making our back sub we get

$\displaystyle -\sqrt{4-x^2}$

3. Originally Posted by dataspot
the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
It's not that obvious at first. But its easy to understand.

$\displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx$

Let $\displaystyle u=4-x^2\implies\,du=-2x\,dx$

Note that in the integral, we have $\displaystyle \int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $\displaystyle -2x\,dx$. To make the $\displaystyle -2x\,dx$ appear, multiply the integral by $\displaystyle \frac{-2}{-2}$ (equivalent to 1)

Thus, we get the following:

$\displaystyle \frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.

Now make the substitutions:

$\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.

Hope this makes sense!

4. Originally Posted by dataspot
the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
the -1/2 is to account for the -2 that was put in front of the x (noticed the -2x?). what they did was multiply the x by -2 and then divide by -2 (hence the -1/2) so as to not change anything. why? because -2x is the derivative of 4 - x^2. why do we care? well, in this set up, it is "easy" to see that we have an expression that is the result of using the chain rule to differentiate a function. basically what happened here is doing substitution without actually saying "let u = 4 - x^2." this is a more clever and concise way of doing it

EDIT: Beaten again...twice...and with better explanations too good job guys

5. Originally Posted by Chris L T521
It's not that obvious at first. But its easy to understand.

$\displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx$

Let $\displaystyle u=4-x^2\implies\,du=-2x\,dx$

Note that in the integral, we have $\displaystyle \int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $\displaystyle -2x\,dx$. To make the $\displaystyle -2x\,dx$ appear, multiply the integral by $\displaystyle \frac{-2}{-2}$ (equivalent to 1)

Thus, we get the following:

$\displaystyle \frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.

Now make the substitutions:

$\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.

Hope this makes sense!
ok...

if the resulting $\displaystyle du$ had equaled $\displaystyle x dx$, then you wouldn't have had to multiply by 1 (rather, $\displaystyle \frac{-2}{-2}$), but since it didn't, you had to even things out.

yeah?

6. Originally Posted by Chris L T521
Now make the substitutions:

$\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.
i'm not sure i could finish the problem from here either. been a while.

7. Originally Posted by dataspot
i'm not sure i could finish the problem from here either. been a while.
from here, we use the power rule: $\displaystyle \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $\displaystyle n \ne -1$

so, $\displaystyle - \frac 12 \int u^{-1/2}~du = - \frac 12 \cdot \frac {u^{1/2}}{1/2} + C = -u^{1/2}$

8. Originally Posted by dataspot
i'm not sure i could finish the problem from here either. been a while.
Use the power rule for integration:

$\displaystyle \int x^n\,dx = \frac1{n + 1}x^{n + 1} + C,\text{ for }n\neq-1$

9. thanks for the help!!!