Question

please, help me find answer: integral (sqrt (36-x^2)xdx). I think, there the first step is: integral (sqrt(36-x^2)dx^2. And what will then?

Answer 1

\[\huge \int\limits \sqrt{36-x^2}xdx\]this it?

Answer 2

yes

Answer 3

Haha :)
Ever heard of u-substitution? :)

Answer 4

yes:) but i don't know how it will be there)

Answer 5

Well, let's play with it a bit, this will help us in the future...
\[\huge \int\limits (\sqrt{36-x^2} \ )\left(-\frac{1}{2} \right)(-2x)dx\]
Essentially the same, right? :)

Answer 6

I just multiplied 2 and divided by 2, negative :)

Answer 7

yes, i understand)

Answer 8

Now, -1/2 is a constant, we can bring it out, right? :)
\[\huge -\frac{1}{2}\int\limits (\sqrt{36-x^2} \ ) (-2x)dx\]

Answer 9

yes, bring it out)

Answer 10

Now, can you see where u-substitution comes in? :)

Answer 11

hm, noo...

Answer 12

Here's a hint. Let
\[\large u = 36-x^2\]
Can you continue?

Answer 13

I will try)

Answer 14

what will v-substitution?

Answer 15

u, v, it doesn't really make a difference. I usually use u first, though

Answer 16

So...\[\large u = 36-x^2\]
Now find \[\large\frac{du}{dx}\]

Answer 17

-2x

Answer 18

precisely :)
\[\large \frac{du}{dx}=-2x\]\[\large du = -2xdx\]
Can you do it now? :)

Answer 19

I understand it) but i don't know about second substitution

Answer 20

No need for that... back to the integral.
\[\huge -\frac{1}{2}\int\limits (\sqrt{36-x^2} \ ) (-2x)dx\]
\[\large u = 36-x^2\]\[\large du=(-2x)dx\]

Replace what can be replaced :)

Answer 21

now i will try)

Answer 22

yees, i could :)

Answer 23

Awesome :)

Answer 24

thank you very much!!:)

Answer 25

Anytime :)

Answer 26

I'm very glad:) i'm newcomer there:)