# Thread: Integration by substitution u^2

1. ## Integration by substitution u^2

I am stuck on this integral, i havent come across $u^2$ before and ive given it a shot but have got stuck where you substitute $\frac{dy}{du}$ in... maybe someone could help

Integrate the following:

$\frac{x}{\sqrt{3-x}} dx; u^2 = 3-x$

2. Originally Posted by darksupernova
I am stuck on this integral, i havent come across $u^2$ before and ive given it a shot but have got stuck where you substitute $\frac{dy}{du}$ in... maybe someone could help

Integrate the following:

$\frac{x}{\sqrt{3-x}} dx; u^2 = 3-x$
If instead you choose u=3-x, du=dx

you get

$\int{\frac{3-u}{u^{\frac{1}{2}}}}du=\int{3u^{-\frac{1}{2}}}du-\int{u^{\frac{1}{2}}}du$

$=\frac{3u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=6u^{\frac{1} {2}}-\frac{2}{3}u^{\frac{3}{2}}+C$

replacing u with 3-x, this is

$6\sqrt{3-x}-\frac{2}{3}\left(\sqrt{3-x}\right)^3+C$

$=6\sqrt{3-x}-\frac{2}{3}(3-x)\sqrt{3-x}+C$

which can be factorised.

3. Originally Posted by darksupernova
I am stuck on this integral, i havent come across $u^2$ before and ive given it a shot but have got stuck where you substitute $\frac{dy}{du}$ in... maybe someone could help

Integrate the following:

$\frac{x}{\sqrt{3-x}} dx; u^2 = 3-x$
u^2 = 3 - x
2u*du = -dx
x = 3 - u^2.
The given integration is
intg[(3 - u^2)(-2u)du/u
intg[2(u^2 - 3)]du.

4. @ Archie Meade the question tells me to use this substitution...

@ sa-ri-ga-ma

I get $\frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}$

Is this correct, how can i simplify this?

I may well have made a mistake as the answer is $-\frac{2}{3}(x+6)\sqrt{(3-x)}$

5. Originally Posted by darksupernova
@ Archie Meade the question tells me to use this substitution...

@ sa-ri-ga-ma

I get $\frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}$

Is this correct, how can i simplify this?

I may well have made a mistake as the answer is $-\frac{2}{3}(x+6)\sqrt{(3-x)}$
'+ C' needs to be included.

To simplify, take out the common factor of $2(3 - x)^{1/2} = 2 \sqrt{3 - x}$ and then simplify.

6. Originally Posted by darksupernova
@ Archie Meade the question tells me to use this substitution...

@ sa-ri-ga-ma

I get $\frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}$

Is this correct, how can i simplify this?

I may well have made a mistake as the answer is $-\frac{2}{3}(x+6)\sqrt{(3-x)}$
$(3- x)^{3/2}= (3- x)^{1+ 1/2}= (3-x)(3- x)^{1/2}$ so you can factor out a $(3- x)^{1/2}$:\
$\frac{2}{3}(3- x)^{3/2}- 6(3- x)^{1/2}= (3- x)^{1/2}(\frac{2}{3}(3- x)- (2/3)(9)$ $-(2/3)(3- x)^{1/2}(9- (3- x)= (3- x)^{1/2}(x+ 6)$.

7. ah thanks HallsofIvy, its stupid the loops i have to jump through, surely seeing i can do the integration would be enough!

Thanks for all the help everyone!