# Integration by Parts for Sec^5x

1. Sep 18, 2007

### rocomath

$$\int\sec^{5}{x}dx$$

i need a hint on this problem, my teacher told us that we prob. can't solve it with the method we've just learned. i'm not sure where to look in the book, i've looked through it but i don't know what is the best method.

hint? plzzz and i'll post all my work :D

i thought of ...

$$\int(\sec^{2}x)^2\sec{x}dx$$

$$\int\frac{\sec^{4}x}{\cos{x}}dx$$

but neither led me anywhere :(

Last edited: Sep 18, 2007
2. Sep 18, 2007

### HallsofIvy

Staff Emeritus
Since sec is 1/cos, that is really cos to an odd power. yes, taking out one to use with dx and converting the remaining even power to sin is one way to do this.
I would write it as
$$\int sec^5 x dx= \int \frac{1}{cos^5 x}dx= \int \frac{1}{cos^6 x} cos x dx$$
Now make that
$$\int \frac{1}{(1-sin^2 x)^3} cos x dx$$
and the substitution u= sin x, du= cos x dx turns it into
$$\int \frac{du}{(1- u^2)^3}$$
Of course, that denominator is of degree 6, but it's already nearly factored for you so "partial fractions" should be a snap!

3. Sep 18, 2007

### dextercioby

When computing $\int \frac{du}{(1-u^2 )^3}$, if you don't like (i don't) partial fractions, you can go for a $u=\tanh t$ substitution.

Daniel.

4. Sep 21, 2007

### rocomath

so i just learned partial fractions yesterday, i will attempt to solve this problem tonight for sure. i appreciate your help, wish me luck!

5. Sep 21, 2007

### rocomath

i'll try that way too, i really want to be able to solve this problem, it's been bugging me for a week. being able to solve it 2 diff. ways will taste so sweet :D

6. Sep 21, 2007

### bob1182006

you could also do that final integral by integration by parts.
but you will need to start from $$\int \frac{1}{1+u^2}$$ and you will arrive at $$\int\frac{1}{(1+u^2)^2}$$ whose answer depends on $$\int\frac{1}{1+u^2}$$, another integration by parts of $$\int\frac{1}{(1+u^2)^2}$$ will give you $$\int\frac{1}{(1+u^2)^3}$$ whose answer depends on $$\frac{1}{(1+u^2)^2}$$.

it's a bit lengthy, like 3-4 lines per integral more or less if you write out every step..

I think even if you do partial fractions you will need to do an integral of 1/(1+u^2)^n where n is 1, 2, and 3 right?

7. Sep 21, 2007

### Gib Z

Dex, I would much rather do partial fractions then solve the resultant $$\int \cosh^4 t dt$$ with two applications of the square exponent to double angle identity = P You must have had a very good teacher do your hyperbolic trig functions to like them this much =]

8. Sep 21, 2007

### rocomath

sigh ... i failed

i did not attempt partial fractions or integration of hyperbolic functions

i'm actually still not on partial fractions in my hw but i was so curious to see where i could get, lol

don't laugh ... :D

http://img215.imageshack.us/img215/2782/0002uj8.jpg [Broken]
http://img215.imageshack.us/img215/5933/0003rk9.jpg [Broken]

i'll attempt it again tomorrow if i finish the section on partial fractions.

Last edited by a moderator: May 3, 2017
9. Sep 22, 2007

### dextercioby

With hyperbolic functions i was my teacher. And i truly dislike partial fractions decompositions when compared to the substitution method.

10. Sep 28, 2007

### rocomath

i still have yet to solve this problem, lol

i've seen the solution through integration by parts, but i want to be able to do it through the partial fractions and hyperbolic method

11. Sep 29, 2007

### Gib Z

Well dex already said how to start for the hyperbolic method, and for partial fractions it should be quite obvious from Halls' post, post 2.

12. Oct 16, 2007

### rocomath

so i got bored and decided to give this problem another try and success! i realized my mistake and it was so stupid.

anyways, if i get bored again i'll try it by integration by parts.

Last edited by a moderator: May 3, 2017
13. Sep 24, 2009

### rockstarpat

So I actually did it by parts. I'm new to the site though and not sure how to put integral signs in the posts though. I tried the diverse symbols latex reference thing but it didn't work for me. Oh well. Anyway... I split up sec^5x into sec^4x*secx, Integrated by parts, pythagorean identitied both of the tan terms that resulted (tan^2x and tan^4x) and that gave me (after distributing everything out) sec^5x again as well as sec^3x and sec^x. I'm pretty sure everyone here has done the last two. No problem. Now sec^5x is replaced with I on both sides and you solve for I to get the integral. I would post it in detail but like I said, I can't get the integral sign to work for me. If anyone wants to tell me how to make it work, I'd appreciate it. The answer I got was 3/4(secx*1/3tan^3x+tanx+(secxtanx+ln|secx+tanx|)/3-1/3ln|secx+tanx|) + C

14. Feb 24, 2010

### texans81

you guys are on the right track but not quite there...let me help :)

I don't know how to make those fancy integral symbols on here either, so I'm gonna use the tilda for it :)

to solve ~sec5x you actually first need to solve ~sec3x. You'll see why when we get there. So lets do that first...

~sec3x = ~sec2xsecx

you want to break it up like this for parts. So remember, when doing integration by parts:

~udv = uv - ~vdu

in this case (although in the original problem they are in reverse order)...
u=secx, dv=sec2x and therefore du=secxtanx and v=tanx

so the problem will now look like this:
~secxsec2x = secxtanx - ~tanx(secxtanx)

~secxsec2x can be rewritten to the original ~sec3x now that we are done determining our parts. And in the second part of the equation, you can multiply the tanx into the secxtanx so you end up with...

~sec3x = secxtanx - ~secxtan2x

so far so good? ok...now use the pythagorean identity to rewrite tan2x in terms of secx, so you now have...

~sec3x = secxtanx - ~secx(sec2x-1)

now distribute the secx and use the rule of adding/subtracting terms in an integral to rewrite this as two integrals:

~sec3x = secxtanx - ~sec3x + ~secx

notice that when you distribute the secx in, you must also distribute the - sign, which is why the last term is now a +

now you'll notice that in the right side of the equation, you actually have the left part of the equation. That's how you know you are on the right track. The next step is to add that portion of the right side of the equation to the left side. After doing that, you're left with:

2~sec3x = secxtanx + ~secx

hopefully everyone follows that. Now, go ahead and simplify the ~secx on the right side of the equation giving you...

2~sec3x = secxtanx + ln|secx + tanx| + c

now to get the original equation, you need to divide both sides by 2.

~sec3x = 1/2secxtanx + 1/2(ln|secx + tanx|) + c

so that is the solution to ~sec3x. Now for sec5x...

again, we'll want to rewrite this to facilitate integration by parts. So here's what we'll end up with.

~sec5x = ~sec3xsec2x

so for the purposes of parts...
u = sec3x, dv = sec2x, and therefore du = 3sec2x(secxtanx) and v = tanx

you use the chain rule to get du incase anyone got lost there...
ok so now that we have our parts, rewrite the left side of the equation back in its original form and use the parts formula
~udv = uv - ~vdu

so we'll have:

~sec5x = sec3xtanx - ~tanx(3sec2x(secxtanx))

that last part looks a little messy, so lets clean it up. You can do this in two steps if you prefer, but I'm just going to do it all at once. We're going to take the tanx and the 3sec2x and multiply then both into the secxtanx so we have one big term there which will be 3sec3xtan2x. So...

~sec5x = sec3xtanx - ~3sec3xtan2x.

Now, as we did before, we're going to use the pythagorean identity to rewrite tan2x in terms of secx.

~sec5x = sec3xtanx - ~3sec3x(sec2x-1)

now distribute the -~3sec3x in and you end up with...

~sec5x = sec3xtanx - ~3sec5x + ~3sec3x

now we're once again at the point where we have the left side of the equation (original problem) present in the right side of the equation. Good stuff. There happens to be a 3 infront of it this time, but that doesn't matter. So, like last time add it to both sides (including the 3 of course) to end up with...

4~sec5x = sec3xtanx + 3~sec3x

remember at the beginning when I said you'd need the solution for ~sec3x? Well now you know why!! Plug it in now and we have...

4~sec5x = sec3xtanx + 3(1/2secxtanx + 1/2(ln|secx + tanx|)

lets go ahead and distribute that 3 now...

4~sec5x = sec3xtanx + 3/2secxtanx + 3/2(ln|secx + tanx|)

now, to get back to our original equation, we need to divide both sides by 4.

~sec5x = 1/4sec3xtanx + 3/8secxtanx + 3/8(ln|secx + tanx|) + c

and there you have it!!

15. Feb 24, 2010

### texans81

now here's a trick or shortcut for doing an anti-derivative of any secx function with an odd power...

~secnx = 1/(n-1)[sec(n-2)xtanx + 1/(n-1)[(n-2)~sec(n-2)x]

that n stuff can be confusing looking so here's a few examples

~sec3x
n = 3
(n-1) = 2
(n-2) = 1

so ~sec3x = 1/2sec1xtanx + 1/2[1~sec1x]

in this case, the 1's aren't necessary so you end up with
~sec3x = 1/2secxtanx + 1/2~secx + c

you of course can simplify the ~secx to be (ln|secx + tanx| + c) so the final answer would be

~sec3x = 1/2secxtanx + 1/2(ln|secx + tanx|) + c

as you get higher and higher in powers, the final answer after you've done all of the simplification is going to get messier and messier, so be forewarned. Lets apply this shortcut to sec5x. so...

~sec5x
n = 5
(n-1) = 4
(n-2) = 3

~sec5x 1/4sec3xtanx + 1/4(3~sec3x)

here's where it starts to get a little messy if you want a final answer...you'll need to plug in the solution you got previously for ~sec3x and then distribute the 3 and then 1/4 into it.

If you go ahead and do that...you'll end up with
~sec5x = 1/4sec3xtanx + 3/8secxtanx + 3/8(ln|secx + tanx|) + c

Basically, if you want to get a final answer with the integrals all simplified for any given odd power of secx, you'll have to know the answer for the previous odd power to do it. So to do sec11x you'll need the answer to sec9x. To do sec9x, you need sec7x and so on, and you'll just plug that answer in to the shortcut formula as I did above to get your final answer. They get reaaaaaaaaally long the higher up you go in powers. Just to illustrate what I mean, here's the final answer for ~sec11x...

~sec11x = 1/10sec9xtanx + 9/80sec7xtanx + 63/480sec5xtanx + 315/1920sec3xtanx + 315/1280secxtanx + 315/1280(ln|secx + tanx|) + c

could I reduce those fractions? yes I could...but after doing all the work to get that, fraction reducing is just not real high on my list

you'll notice two things if you go through and do all of these expansions by hand. First...each time you go up a power (1,3,5,7,9,etc...) you end up with one more "term" than in the previous solution. For example...

~secx = ln|secx + tanx| + c (two terms if you count "c")
~sec3x = 1/2secxtanx + 1/2(ln|secx + tanx|) + c (3 terms including "c")
~sec5x = 1/4sec3xtanx + 3/8secxtanx + 3/8(ln|secx + tanx|) + c (4 terms including "c")

and then if we look at ~sec11x that I just showed you, following this pattern it should have 7 terms including "c" and you can see here that indeed it does.
~sec11x = 1/10sec9xtanx + 9/80sec7xtanx + 63/480sec5xtanx + 315/1920sec3xtanx + 315/1280secxtanx + 315/1280(ln|secx + tanx|) + c

the second thing you'll notice is that no matter what power you are doing, your last term before "c" will ALWAYS be a/b(ln|secx + tanx|)

hope you guys find this info useful!!

16. Apr 9, 2011

### Faye

I don't know if I am correct, but I used Trigonometric Transformation:

Solution

Please correct me if I am wrong. This question also came up during my finals, but since I was unable to solve it using trig trans (due to time pressure) I actually went for integration by parts ;_; which is fairly difficult >_<

Last edited: Apr 9, 2011
17. Apr 10, 2011

### vela

Staff Emeritus
No, that's not correct. The derivative of tan x is sec2 x, not sec x, so you can't integrate each term the way you did.

18. Jun 19, 2011