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Evaluating the integral, correct? 
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#73
Jan2008, 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



#74
Jan2008, 12:49 AM

P: 62

woo now i know 6 out of the 18 problems i have to do are correct thanks to you



#75
Jan2008, 12:51 AM

P: 1,754




#76
Jan2008, 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 


#77
Jan2008, 12:57 AM

P: 1,754

[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. 


#78
Jan2008, 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 


#80
Jan2008, 01:08 AM

P: 62

woot lol



#81
Jan2008, 01:13 AM

P: 62

im goin to let my brain cool down for the night, t2ul and have a good night.



#82
Jan2008, 01:13 AM

P: 1,754




#83
Jan2008, 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 


#84
Jan2008, 01:49 PM

P: 1,754

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 


#85
Jan2008, 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 


#87
Jan2008, 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 (1sin^2x), then parts? 


#88
Jan2008, 02:09 PM

P: 1,754




#89
Jan2008, 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 


#90
Jan2008, 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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