Question

Use a change of variables to find the following indefinite integral. Check your work by differentiation

Answer 1

\[\int\limits_{?}^{?}\frac{ 3x^3 }{ ? \sqrt{5-6x^4}}dx\]

Answer 2

omit/ignore the question marks

Answer 3

let A = 3x^2

Answer 4

hey e .mork u speak chinese or japanese?

Answer 5

Japanese

Answer 6

A nice substitution would be:
\[\beta = 5-6x^4 \implies d \beta = -24x^3dx \implies \frac{d \beta}{-8} = 3x^3 dx\]
This gives:
\[\rightarrow \frac{-1}{8} \int\limits \frac{d \beta}{\sqrt{\beta}}=\frac{-1}{8} \int\limits \beta ^ {- \frac{1}{2}}d \beta\]
I'll leave the rest to you. Don't forget to plug what beta is in terms of x after you integrate.

Answer 7

either way.

Answer 8

cool, i speak chinese... btw all the extra stuff arrountd the problem is confusing me , glad you dont need help anymore...lol

Answer 9

I do need help....

Answer 10

oh well could u re-write the question? maybe i can help you.. idk though

Answer 11

@malevolence19 is it -1/8 + C?

Answer 12

-1/4X^(1/2) +C?

Answer 13

Well when you integrate you'll have:
\[\frac{-1}{4} \beta ^{\frac{1}{2}} + C = \frac{-1}{4} \sqrt{5-6x^4} + C\]

Answer 14

When you differentiate, don't forget the chain rule!

Answer 15

Okay Thanks soo much!!

Answer 16

good job malevolence19... Lol of course i geet the question at the end...lol

Answer 17

good job malevolence19... Lol of course i geet the question at the end...lol

Answer 18

No offense drizzy, but what are you even talking about?

Answer 19

nevermind..