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Since the powers are both you can strip one tangent or one secant and use the trig identity \[\tan^2x+1=\sec^2x\]
okay!! so what happens next? :/
You can make a u substitution, have you tried anything?
so u = tan^5(x)sec^3(x) and dv = dx ?
ohhh okay so what would be my next step? :/ sorry, integrals confuse me a lot :(
You'll need to remember that (secx)' = secxtanx
You should try messing with it yourself @iheartfood
lets do u = secx then du = secxtanxdx better to do this \[\int\limits \tan^4xsec^2x secxtanx dx\]
Save some time this way
Yea astrophysics your second way is better.... then invert the relationship you gave to put it all in terms of sec(x)
Yup, it just goes to show, there's many ways to approach these :P
\[ \int\tan^5(x)\sec^3(x)\,dx\\ \tan^2(x)+1=\sec^2(x)\\ \int\left(\sec^2(x)-1\right)^2(x)\sec^3(x)\tan(x)\,dx\\ u=\sec(x)\\ du=\sec(x)\tan(x)\,dx\\ \int\left(u^2-1\right)u^2\,du \] A bit late lol.
ooh okay haha thank you both!! and so would i sub in now?
\[ \int tan^4(x)sec^2(x) sec(x)tan(x)dx = \int tan^4(x)sec^2(x) d(sec(x)= \int (1-sec^2(x))^2sec^2(x) d(sec(x) \] Then change variables to u=sec(x) \[ \int (1-u^2)^2u^2du\]
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
You can integrate it now from \[\int\limits (u^2-1)^2 u^2 du\]
Either integration by parts or multiply the whole thing out.
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 ?
Hard to tell what you're doing \[\int\limits (u^2-1)^2u^2 du \implies \int\limits (u^4-2u^2+1)u^2 du\]
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
Equations *button* form :D
I gave you one, sorry thomas took one from you so everyone would have one haha :P
ohhh :O wait so do i distribute next?
Yes, there still is the u^2
Oh wow I didnt see Thomas had beaten me to the punch he does deserve the medal there... I was posting in another problem
so then i get integral of u^6 - 2u^4 + u^2 du ?
okay! and so i integrate and get this? |dw:1442282862540:dw|
You drop the integral sign and du when you integrate, now you can plug back your u substitution and add a +C
ohh okay i see so i would end up with this? |dw:1442282938959:dw|
ooh yay! and so I'm done? :O
yay!! thank you so much!!:D
Np @thomas5267 and @PlasmaFuzer did great work to, collaborative, which makes it funner!
So Mr Astrophysics has beaten me has beaten PlasmaFuzer? An ordering on OpenStudy lol?
Haha makes sense @thomas5267 sorry for trying to steal your thunder im bouncing around in too many questions I guess
Definitely agree @Astrophysics... Its also more fun to do harder problems :D
Sometimes it is really fun to race with others and spit out the correct answer fastest!
Yea totally agree it adds to the excitement since many questions can be fairly dull
Most question here are quite dull. However I am not that good to answer questions on Math StackExchange.
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.