Question

Integral Brainfart

Answer 1

\[\int\limits_{}^{}e ^{4x}\sqrt{1-e ^{2x}}\]

Answer 2

Not sure how to solve this integral

Answer 3

put e^2x = u
du = 2e^2x

Answer 4

to get:
\[1/2\int\limits_{}^{}u ^{2}\sqrt{1-u}du\] ?

Answer 5

yeah, something like that.

Answer 6

or actually is it just u in front of the square root

Answer 7

since e^4x = (e^2x)(e^2x)

Answer 8

and e^2x is contained in du

Answer 9

\[1/2∫u \sqrt{1-u}du\]

Answer 10

Like that

Answer 11

yes, that is correct.

Answer 12

And then is in integration by parts to get the rest?

Answer 13

sooo,
u = u
du = du
dv = sqrt(1-u)
v = ?

Answer 14

Hmm, how would I integrate dv?

Answer 15

nvm, u-sub

Answer 16

\[\int\limits e^{4x} \sqrt{1-e^{2x}}dx, u=1-e^{2x}, du= -2e^{2x}dx, \frac{du}{-2e^{2x}}=dx\]
\[\int\limits \frac{e^{4x} \sqrt{u}}{-2e^{2x}}du \rightarrow \frac{1}{2} \int\limits e^{2x} \sqrt{u}du\]
now since:
\[u= 1-e^{2x}, e^{2x}= 1-u\]
Now back substitute 1-u:
\[\int\limits (1-u) \sqrt{u }du \rightarrow \int\limits (\sqrt{u}-u^{\frac{3}{2}}) du\]
\[=\frac{2}{3}u^\frac{3}{2}- \frac{2}{5}u^\frac{5}{2}+c\]
since: \[u=1-e^{2x}\]
\[=\frac{2}{3}(1-e^{2x})^\frac{3}{2}- \frac{2}{5}(1-e^{2x})^\frac{5}{2}+c\]

Answer 17

come help me nadeem!!

Answer 18

Thanks nadeen I'll use this to check my work! :D

Answer 19

typo multiply the final answer my (-1/2)

Answer 20

also after I back substituted e^2x=1-u I forgot to include the (-1/2) constant, sorry

Answer 21

You final answer should be:
\[\frac{1}{5}(1-e^{2x})^\frac{5}{2}-\frac{1}{3}(1-e^{2x})^\frac{3}{2}+c\]