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Evaluating the integral, correct?

by Zack88
Tags: correct, evaluating, integral
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Zack88
#73
Jan20-08, 12:47 AM
P: 62
lol i was looking at it and then i tilted my head and was like yep i dont know how that happened
Zack88
#74
Jan20-08, 12:49 AM
P: 62
woo now i know 6 out of the 18 problems i have to do are correct thanks to you
rocomath
#75
Jan20-08, 12:51 AM
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P: 1,755
Quote Quote by Zack88 View Post
woo now i know 6 out of the 18 problems i have to do are correct thanks to you
I'm getting sleepy, if you wanna do a couple more better start asking :-]
Zack88
#76
Jan20-08, 12:54 AM
P: 62
me too me too

[integral] sec^6 x dx

should i do

[integral] (sec^2)^3

or

start doing parts

u= sec^2 x dv = sec^4 x
rocomath
#77
Jan20-08, 12:57 AM
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P: 1,755
[tex]\int\sec^6 xdx[/tex]

[tex]\int\sec^4 x \sec^2 x dx[/tex]

[tex]\int(\sec^2 x)^2 \sec^2 xdx[/tex]

When you trig identities raised to powers, break it up till you see something.
Zack88
#78
Jan20-08, 01:03 AM
P: 62
[integral] (tan^2 x +1) sec^2 x dx

u = tan x
du = sec^2 x

[integral] (u^2 =1)^2 du

[integral]u^4 + 2u^2 + 1 du

tan^5 x / 5 + 2tan^3 x / 3 + tan x + c
rocomath
#79
Jan20-08, 01:06 AM
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P: 1,755
Quote Quote by Zack88 View Post
[integral] (tan^2 x +1) sec^2 x dx

u = tan x
du = sec^2 x

[integral] (u^2 =1)^2 du

[integral]u^4 + 2u^2 + 1 du

tan^5 x / 5 + 2tan^3 x / 3 + tan x + c
Good!!!
Zack88
#80
Jan20-08, 01:08 AM
P: 62
woot lol
Zack88
#81
Jan20-08, 01:13 AM
P: 62
im goin to let my brain cool down for the night, t2ul and have a good night.
rocomath
#82
Jan20-08, 01:13 AM
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P: 1,755
Quote Quote by Zack88 View Post
woot lol
Alright, I'm going to sleep! You still have Sunday and Monday to finish all 18 problems!!!
Zack88
#83
Jan20-08, 01:40 PM
P: 62
woo im back, v.v

[integral] sin^5 x cos^3 x

i was wanting to know if i should break it up like

[integral] sin^4 x sin x cos^2 x cos x
rocomath
#84
Jan20-08, 01:49 PM
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P: 1,755
Look in your book, there should be a suggestion on how to tackle even/odd powers of sines and cosines.

Hint: Leave sin^5 x alone, mess around with cos^3 x
Zack88
#85
Jan20-08, 02:00 PM
P: 62
[integral] sin^5 x cos^2 x cos x
[integral] sin^5 x (1- sin^2x) cos x

u= sin
du= cos

[integral] u^5(1- u^2) du
[integral] u^5 - u^7 du

1/6 sin^6 x - 1/8 sin^8 x + c
rocomath
#86
Jan20-08, 02:01 PM
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Quote Quote by Zack88 View Post
[integral] sin^5 x cos^2 x cos x
sin^5 x (1- sin^2x) cos x
u= sin
du= cos

[integral] u^5(1- u^2) du
u^5 - u^7 du

1/6 sin^6 x - 1/8 sin^8 x + c
Good!
Zack88
#87
Jan20-08, 02:06 PM
P: 62
woo, for

[integral] x cos^2 x dx

do i want to use

cos^2 x = (1 + cos^2 x) / 2, then parts

or

[integral] x (1-sin^2x), then parts?
rocomath
#88
Jan20-08, 02:09 PM
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P: 1,755
Quote Quote by Zack88 View Post
woo, for

[integral] x cos^2 x dx

do i want to use

cos^2 x = (1 + cos^2 x) / 2, then parts

or

[integral] x (1-sin^2x), then parts?
Try a method! Come test day, you gotta just go at it :-] You've handled harder problems than this, I'm sure you can do this with ease.
Zack88
#89
Jan20-08, 03:18 PM
P: 62
sorry had a fire drill and then went to go eat

im goin to do the (cos^2 x)/ 2 and then the chain rule
Zack88
#90
Jan20-08, 03:34 PM
P: 62
[integral] x cox^2 x dx

u= x
du = dx

dv= cos^2
v= (sin(2x)) / 4 + x/2

x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du

x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c


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