Question

Practising for the substitution rule. The question is as follows:\[\int {x \over \sqrt{1 - 4x^2}}dx \]I need you guys to check each of my step one-by-one. Thank you.

Answer 1

Okay, so first I take \(u = \sqrt{1 - 4x^2}\)

Answer 2

yes

Answer 3

it's better to take u = 1 - 4x^2 actually

Answer 4

Then \(du = -8xdx\)

Answer 5

no..wait...remove square root

Answer 6

Sorry...

Answer 7

proceed

Answer 8

Ok, then\[\int x(1 - 4x^2)^{-1 \over 2} dx \]

Answer 9

mmhmm

Answer 10

Oh, and \(xdx = {-1 \over 8}du\)

Answer 11

Right?

Answer 12

yes...

Answer 13

\[\int u^{-1 \over 2} xdx \]

Answer 14

mmhmm

Answer 15

substitute the xdx

Answer 16

\[\int u^{-1 \over 2} {-1 \over 8}du \]

Answer 17

go on...

Answer 18

\[{-1 \over 8}\int u^{-1 \over 2} du \]

Answer 19

yes

Answer 20

good

Answer 21

\[{-1 \over 8}u^{1 \over 2} + c \]

Answer 22

-1/4*

Answer 23

yepp

Answer 24

And then you substitute \(u\) back.

Answer 25

\[{-1\over 4}(1 - 4x^2)^{1 \over 2} +c\]

Answer 26

roght

Answer 27

let me show you what happens if you let \(u = \sqrt{1-4x^2}\) (because im bored)

\[\implies \int \frac{xdx}{\sqrt{1-4x^2}}\]
let \[a = \sqrt{1-4x^2}\]
\[a^2 = 1 - 4x^2\]
\[2ada = -8xdx\]
\[-\frac{2ada}{8} = xdx\]
\[-\frac{ada}{4} = xdx\]
so if you substitute...
\[\implies \int (\frac 1a) (-\frac{ada}4)\]
\[\implies \int -\frac{da}4\]
\[\implies -\frac a4 + C\]
sub back...
\[\implies -\frac{\sqrt{1 - 4x^2}}4 + c\]
same shiz

Answer 28

this is actually a method called algebraic substitution (simpler variation) that is very similar to u-sub in this scenario..bu usually they are very different

Answer 29

lol, same shiz.

Answer 30

it's the only expression i have left that hasn't been commercialized...now that <tips hat> is gone