Question

Integrate "square root of x times sec^2(1+x^3/2)dx? Please help or give hints?

Answer 1

Try \[

y=1+x^{3/2}\quad dy=\frac 32 \sqrt x\;dx
\]

Answer 2

Also remember \(d(\tan x)=(\sec^2 x)\;dx\)

Answer 3

Okay, so the way our professor taught us, we need to use the substitution method. So if I find u= 1+ x^3/2 and then du=3/2 sqrt x...I'm kind of confused what to do next? :/

Answer 4

Do I just start taking antiderivative of sqrtxsec^2(1+x^3/2)?

Answer 5

well

Answer 6

\[
\sec^2(1+x^3/2) = \sec(u)
\]

Answer 7

Start with that.

Answer 8

Okay, I remember sec(u)=sec(u)tan(u)

Answer 9

\[
\sqrt x \;dx= \frac 23 \left(\frac 32 \sqrt x\;dx\right) = \frac 23 du
\]

Answer 10

=\[\int\limits \sqrt(x)\sec^2(1+x ^{3/2})dx\]
=\[u=1+x^{3/2}\]
=\[du=3/2\sqrt{x} dx \]
So,
=\[\int\limits 2/3(3/2\sqrt(x)dx) \sec^2(1+x ^{3/2})\]
=\[\int\limits 2/3 du \sec^2(u)\]
=\[=2/3\tan(1+x^{3/2}) + c\]

Answer 11

okay that's what I have so far....And I know it's terribly wrong... :/ Confused..

Answer 12

Question - where is the 2/3 coming from in the 4th step @aimee19

Answer 13

oh, okay so this is what I was thinking. so you know how I got

\[du=3/2\sqrt{x}dx\]
So then I have to find a way to substitute du into the original equation, right? but there's no \[3/2\sqrt{x}dx \] in the original equation. However

Answer 14

\[\sqrt{x}=\frac{ 2 }{ 3 }(\frac{ 3 }{ 2 }\sqrt{x})\]

Answer 15

so I can rewrite the original equation which was \[\int\limits \sqrt(x)\sec^2(1 + x^\frac{ 3 }{ 2 }) dx \]
=
\[\int\limits \frac{ 2 }{ 3 }(\frac{ 3 }{ 2 }\sqrt(x) )\sec^2( 1 + x^\frac{ 3 }{ 2 })\]

Answer 16

so then I can now substitute du in.

Answer 17

...but I'm not sure I'm even doing the right thing. I was trying to do the substitute method to integrate the problem.

Answer 18

Yeah you only use sub. when they give you a specific range to integrate on

Answer 19

and the 3/2 and 2/3 cancel each other out

Answer 20

you cant forget to leave out the dx, which prevents that.

Answer 21

so the substitute method shouldn't be used to solve this?

Answer 22

yes you can still use substitution but you're just left with a constant "c" at the end of the final result. this is the best way to proceed, keep going until you have the right answer

Answer 23

I'm not sure if this is right, but if I continue this is what I get:

\[\int\limits 2/3 du sex^2(u) \]
=\[2/3 \tan(u) + c\]
=\[=2/3 \tan (1 + x^\frac{ 3 }{ 2 }) + c \]

Answer 24

is that correct?

Answer 25

standby

Answer 26

alright (:

Answer 27

I'm pretty sure I messed up somewhere...

Answer 28

yes, i think so. i'm still working on it

Answer 29

i dont think it can be integrated, but i'm still looking at it

Answer 30

my mistake! i left something out give me another 5 minutes

Answer 31

no rush! I'm basically spending the whole night working on as much review problems as I can (: I appreciate you taking the time to help!

Answer 32

your method was correct and the answer is definitely 2/3tan(1 + x^3/2) + c

Answer 33

well done!

Answer 34

wait...I was actually right??!

Answer 35

that's like a first for me lol >.< Thank you so much!!

Answer 36

if you differentiate this result you'll get the original integral!

Answer 37

ah, I didn't think of checking that way. I'll use that in the future to check my answers. thanks for your help! (:

Answer 38

no problem you're welcome