Question

anyone good at calculus, Please give me a hand \[\int\limits_{}^{} e ^{xsec(e(x))^3} dx\]

Answer 1

\[\int\limits \frac{ (1-x^2)^\frac{ 2 }{ 3 } }{ x^6 } dx\]

Answer 2

\[\int\limits e^{xsec(e^x)^3} dx\]

Answer 3

please someone help me out with those, two integrals

Answer 4

What chapter is this?? U substitution?

Answer 5

integral technics

Answer 6

Have you learned about u sub yet?

Answer 7

yeah

Answer 8

U sub, by parts, partial fractions

Answer 9

could you help me out ?

Answer 10

IDk how to start,

Answer 11

@dan815 I think these integrals are too advanced for me. And too advanced for wolfram. You want to take a look?

Answer 12

Well at least one is too hard for wolfram. But both cannot be expressed as elementary functions.

Answer 13

damn those integrals are trick, indeed..

Answer 14

This is the exact way the problems look?

Answer 15

I think Im going to fail my midleterm

Answer 16

i believe it is integration by parts, that or you would have to do a fraction decompisition...Its been a while since calc two so i cant help you out there. sorry

Answer 17

:/ ok ...

Answer 18

oh ya sec is never good on e

Answer 19

what book are you using and what problem is it. Sometime you can find solutions

Answer 20

the best u can do is expand sec as a power series and rewrite a product log definition of the integral

Answer 21

I wonder if the second one is actually suppose to be :
\[\int\limits_{}^{}e^x \sec^3(e^x) dx \text{ if so that is tons doable }\]

Answer 22

otherwise u have to resort to nonelementary functions

Answer 23

yeah i believe so is sec ^3

Answer 24

and the sec is not part of the exponent right?

Answer 25

let me take a look

Answer 26

oh yeah lucky for me it is not

Answer 27

\[\int\limits_{}^{}e^x \sec^3(e^x) dx \\ \text{ sub: } u=e^x \\ du=e^x dx \\ \text{ then apply integration by parts } \]

Answer 28

sounds fair

Answer 29

what about the first one ?

Answer 30

Will you confirm that is actually:
\[\int\limits_{}^{}\frac{(1-x^2)^\frac{2}{3}}{x^6} dx\]

Answer 31

100 sure

Answer 32

its a partial fraction decompisistion according to wolfram
http://www.wolframalpha.com/input/?i=integral+%28%281-x%5E2%29%5E%283%2F2%29%29%2Fx%5E6

Answer 33

if only it was 3/2 :(

Answer 34

oh lol

Answer 35

my bad

Answer 36

I cannot do this because I'm not familiar with hypergeometric function . lol.

Answer 37

do me a favor and punch your teacher in the face