Question

With regards to Volume of Revolution for f(x) = sec^2(x), I need to integrate sec^4(x). Which is best approach - substitution or by parts?

Answer 1

once you have sec^n(x), where n > 3, you'll have to use parts

Answer 2

but we already had reduction formula, so it isn't that bad

Answer 3

Thanks but the reduction formula isn't part of this module's content so I have to use ither parts or substitution

Answer 4

Is the trick to treat it as (sec^2(x))(tan^2(x) +1)?

Answer 5

substitution won't work, because one term isn't the derivative of the others.

So you'll have to break it up as sec^2(x) sec^2(x), use part.

u = sec^2(x), dv = sec^2(x)

Answer 6

Ok so I get \[\sec ^{2}x tanx - \frac{ 2\tan ^{3}x }{3}\]
This seems a little complicated compared to the rest of the examples. Going back to the Volume of Revolution, am I correct in saying that to do \[\pi \int\limits_{b}^{a} (\sec^{2}x)^{2}\] I need to to find \[\pi \int\limits_{b}^{a} \sec^{4}x\]?

Answer 7

parts

Answer 8

I used parts to get the top equation

Answer 9

so you did
u = sec^2x dv = sec^2(x)
du = 2sec^2x tanx u = tan^2(x)

?

Answer 10

u = tan(x) , typo

Answer 11

Yes

Answer 12

Must've gone wrong or need further simplifying

Answer 13

sec^2(x) tan(x) - 2∫tan(x) sec^2(x) dx

is what you should have had

Answer 14

I had sec^2(x) tan(x) - 2∫tan^2(x) sec^2(x) dx

Answer 15

sec(x) sec(x) tan(x)

u = sec(x)
du = sec(x) tan(x)
∫u du = (1/2) u^2 = (1/2) sec^2(u)

so 2 [(1/2) sec^2(x)] = sec^2(x)

Answer 16

it's tan(x) sec^2(x), not tan^2(x) sec^2(x)

Answer 17

sec^2(x) tan(x) - sec^2(x) is the final answer

Answer 18

or sec^2(x) (tan(x) - 1)

Answer 19

ahhhh..... I think I did something wrong O.O

Answer 20

ok, let me back up

u = sec^2(x), dv = sec^2(x)
du = 2 sec(x) sec(x) tan(x), v = tan(x)
so,

sec^2(x) tan(x) - 2 ∫ sec(x) sec(x) tan(x) dx

looks good?

Answer 21

u = sec(x)
du = sec(x) tan(x)

then ∫u du = (1/2) u^2

what the... I did right :/

Answer 22

Yeh ok, I can see that. But is there another way of writing sec^2x in terms of sin, cos or tan so I can input the limits?

Answer 23

ahahah, I did it it wrong :/

should have been -2∫sec^2(x) tan^2(x) dx

XD

Answer 24

so you were right :D

Answer 25

I was just looking back through it and getting confused! :D

Answer 26

so, 2∫sec^2(x) (sec^2(x) - 1) dx

2∫sec^4 dx - 2sec^2(x) dx

add 2 sec^4(x) for both sides, and integrate the last term
3∫sec^4(x) = sec^2(x) tan(x) - 2∫sec^2(x) dx

3 ∫sec^4(x) = sec^2(x) tan(x) - 2tan(x)

= (1/3) sec^2(x) tan(x) - (2/3) tan(x) <---

Answer 27

apparently still wrong. Must have had algebraic mistake somewhere :/

Answer 28

I wonder if I'm accidentally making it too complicated. The full question is: Find the volume of the solid of revolution obtained when the graph of f(x) = sec^2x from -pi/4 to pi/3 is rotated about the x axis

Answer 29

ah found it. Off by a minus sign. Should have been
(1/3) sec^2(x) tan(x) + (2/3) tan(x)
^ plus, not minus

Answer 30

I know the answer from Mathcad and Wolfram Alpha to be 15.072. Before multiplying by pi it should be 4.797

Answer 31

just evaluate (1/3) sec^2(x) tan(x) + (2/3) tan(x) from -pi/4 to pi/3

Answer 32

Ok thanks - nearly there! How do I evaluate a value with sec^2x (or indeed secx) Is it just that it's 1/cosx?

Answer 33

correct

Answer 34

Brilliant thanks - I'll give that a go

Answer 35

and multiply by pi will give the the answer

Answer 36

Got it, fantastic! Thanks very much for your help!

Answer 37

yw