Question

Calculate the following integral. $(z+2)(rt(1-z))dz

Answer 1

you mean \[\int \dfrac{z+2}{\sqrt{1-z}}dz\]

Answer 2

or, \[\int {(z+2)}{\sqrt{1-z}}dz \]

Answer 3

This is a problem from a quiz concerning the section on u-substitution. I got it wrong, so maybe you guys can help me figure it out!
@hartnn - no, I do not, I mean for the second term to be also on top.

Answer 4

The second one is what I mean.

Answer 5

right, so put \(\sqrt{1-z}=u \\ then, z=1-u^2\)
dz=..?

Answer 6

Wouldn't it be simpler to set u=1-z?
du=(1/(2((1-z)^(-1/2)))(-1)dz

Answer 7

simpler ? no....
du=(1/(2((1-z)^(-1/2)))(-1)dz itself looks complicated compared to
dz = -2u du

Answer 8

Quite true...

Answer 9

so, your new integral after substitution will be.... ?

Answer 10

-2$(3u-u^3)(rt(u))du?

Answer 11

how come root sign come again! ?
that substitution was done to eliminate the \(\sqrt . \) sign...

Answer 12

Oh, whoops! No root. But is the rest correct?

Answer 13

-2u (3-u^2)u du
yes, thats correct :)

Answer 14

u know how to proceed ?

Answer 15

Yep! Thank you!

Answer 16

welcome ^_^