Question

evaluate the integral.

Answer 1

Post away

Answer 2

\[\int\limits \frac {1}{\sqrt{-x^2-4x}}\]

Answer 3

it says complete the square if necessary :).

Answer 4

Oh ok.. that would work

Answer 5

-x^2-4x=-(x^2+4x+4)+4=-(x+2)^2+4

Answer 6

Did you follow how I completed the square? The inside has a +4, because with the - sign outside it would become -4, and cancel out with the +4 on the outside

Answer 7

So your integral becomes:
Int(1/sqrt(4-(x+2)^2),x)

Answer 8

Oic :).

Answer 9

That looks, ugly, I'll rewrite it:
\[\int \frac{1}{\sqrt{4-(x+2)^2} }dx\]
Substitute:
u=x+2, du=dx, so the integral becomes:
\[\int \frac{1}{\sqrt{4-(u)^2} }du\]

Answer 10

arcsin :D.

Answer 11

arcsin (x+2)/2?

Answer 12

+c

Answer 13

\[\int \frac{1}{\sqrt{2^2-u^2} }dx= \]Sin^-1, yea you're right lol you got it before I managed to figure out how to type ArcSin here

Answer 14

:)

Answer 15

Thanks sir :).

Answer 16

the sub you want is\[x+2=2\sin u\]just to be clear

Answer 17

(different u that above)

Answer 18

than*

Answer 19

But this way it falls nicely into ArcSin integral

Answer 20

the substitution I prescribed would prove that formula, but yeah it's arcsin

Answer 21

Ohhh

Answer 22

\[x+2=2\sin u\implies dx=2\cos udu\]the integral is\[\int{2\cos udu\over\sqrt{2^2-(2\sin u)^2}}=\int{}\frac22\frac{\cos udu}{\sqrt{1-\sin^2 u}}\]remembering\[1-\sin^2 u=\cos^2u\]we get\[\int(1)du=u+C\]and since we know that\[x+2=2\sin u\implies u=\sin^{-1}(\frac{x+2}2)\]you get the same result :)

Answer 23

:)

Answer 24

1.013, -5.918

and

-1.013,9.059

Answer 25

you didn't mention it was a definite integral, so I'm not seeing where those numbers are coming from

Answer 26

wait.... This is the wrong question for that answer sorry :P.

Answer 27

lol

Answer 28

I thought this was the rel min/maxquestion sorry! :P.