- anonymous

find the integral of tan^5(x)sec^3(x) dx
how do I do this? thank you!

- jamiebookeater

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- anonymous

|dw:1442281422437:dw|

- Astrophysics

Since the powers are both you can strip one tangent or one secant and use the trig identity \[\tan^2x+1=\sec^2x\]

- anonymous

okay!! so what happens next? :/

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## More answers

- Astrophysics

You can make a u substitution, have you tried anything?

- anonymous

so u = tan^5(x)sec^3(x) and dv = dx ?

- anonymous

ohhh okay so what would be my next step? :/ sorry, integrals confuse me a lot :(

- beginnersmind

You'll need to remember that
(secx)' = secxtanx

- Astrophysics

You should try messing with it yourself @iheartfood

- Astrophysics

oh wait

- Astrophysics

lets do u = secx then du = secxtanxdx better to do this \[\int\limits \tan^4xsec^2x secxtanx dx\]

- Astrophysics

Save some time this way

- anonymous

Yea astrophysics your second way is better.... then invert the relationship you gave to put it all in terms of sec(x)

- Astrophysics

Yup, it just goes to show, there's many ways to approach these :P

- thomas5267

\[
\int\tan^5(x)\sec^3(x)\,dx\\
\tan^2(x)+1=\sec^2(x)\\
\int\left(\sec^2(x)-1\right)^2(x)\sec^3(x)\tan(x)\,dx\\
u=\sec(x)\\
du=\sec(x)\tan(x)\,dx\\
\int\left(u^2-1\right)u^2\,du
\]
A bit late lol.

- Astrophysics

Hehe

- anonymous

ooh okay haha thank you both!! and so would i sub in now?

- anonymous

\[ \int tan^4(x)sec^2(x) sec(x)tan(x)dx = \int tan^4(x)sec^2(x) d(sec(x)= \int (1-sec^2(x))^2sec^2(x) d(sec(x) \] Then change variables to u=sec(x) \[ \int (1-u^2)^2u^2du\]

- anonymous

- anonymous

Darn it got cut off... well the only change between the second and third step was subbing in the fact that tan^4=(tan^2)^2 and then tan^2=1-sec^2

- Astrophysics

You can integrate it now from \[\int\limits (u^2-1)^2 u^2 du\]

- anonymous

^ :D

- thomas5267

Either integration by parts or multiply the whole thing out.

- anonymous

so i get integral of (u^4-1)(u^2)du --> integral of u^6-u^2 du
so then i get ((u^7)/7) - ((u^3)/3) du ?

- Astrophysics

Hard to tell what you're doing \[\int\limits (u^2-1)^2u^2 du \implies \int\limits (u^4-2u^2+1)u^2 du\]

- anonymous

Yes the equations form is by far superior... btw can I get one medal pretty please though astro deserves the rest since he is finishing it off

- anonymous

Equations *button* form :D

- Astrophysics

I gave you one, sorry thomas took one from you so everyone would have one haha :P

- anonymous

TY <3

- anonymous

lol

- anonymous

ohhh :O
wait so do i distribute next?

- Astrophysics

Yes, there still is the u^2

- anonymous

Oh wow I didnt see Thomas had beaten me to the punch he does deserve the medal there... I was posting in another problem

- anonymous

Whoopsie :/

- anonymous

so then i get integral of u^6 - 2u^4 + u^2 du ?

- Astrophysics

Yes

- anonymous

okay! and so i integrate and get this?
|dw:1442282862540:dw|

- Astrophysics

You drop the integral sign and du when you integrate, now you can plug back your u substitution and add a +C

- anonymous

ohh okay i see so i would end up with this? |dw:1442282938959:dw|

- Astrophysics

\[\large \checkmark\]

- anonymous

ooh yay! and so I'm done? :O

- Astrophysics

Yes

- anonymous

yay!! thank you so much!!:D

- Astrophysics

Np @thomas5267 and @PlasmaFuzer did great work to, collaborative, which makes it funner!

- thomas5267

So Mr Astrophysics has beaten me has beaten PlasmaFuzer? An ordering on OpenStudy lol?

- anonymous

Haha makes sense @thomas5267 sorry for trying to steal your thunder im bouncing around in too many questions I guess

- anonymous

Definitely agree @Astrophysics... Its also more fun to do harder problems :D

- thomas5267

Sometimes it is really fun to race with others and spit out the correct answer fastest!

- Astrophysics

Hehe agreed!

- anonymous

Yea totally agree it adds to the excitement since many questions can be fairly dull

- thomas5267

Most question here are quite dull. However I am not that good to answer questions on Math StackExchange.

- Astrophysics

There can be good questions, it's just hard to find the ones you're interested in, and most other sites I've encountered aren't as kind as the people here, very discouraging at times.

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