Can anyone help me with integration by parts? - 125 4t tuning
I need more help with this integral (with integration by parts). I have the answer to their suite. I could really help, especially with the first! Thanks in advance!
1. The integral of: x ^ 4 (lnx) ^ 2 dx (2 with upper and lower limit of 1)
Answer: (32 / 5) (LN2) ^ 2 - (64/25) ln2 + 62/125
2. The full DT: arctan4t
Answer: arctan 4t t - (1 / 8) (1 + ln 16t ^ 2) + 2
Tuesday, December 8, 2009
125 4t Tuning Can Anyone Help Me With Integration By Parts?
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2 comments:
Now the integral is
1 / 4 ∫ sec ^ 2 x dx (x)
Integration of parts
u = x: du = dx
dv = s ^ 2 (x): v = tanx
1 / 4 ∫ x sec ^ 2 (x) dx = uv - ∫ VDU
1 / 4 [x tanx - ∫ tanx dx]
=> 1 / 4 [x tanx + ln (cos)
If tanx = 4T: sin x / cos x = 4T: sqrt (1-cos ^ 2 (x)] / cos x = 4t
16t ^ 2 cos ^ 2 (x) = 1 - cos ^ 2 (x): 16t ^ 2 cos ^ 2 (x) ^ 2 + cos (x) = 1
cos ^ 2 (x) [16t ^ 2 + 1] = 1
cos ^ 2 (x) = 1 / (1 + 16t ^ 2)
cos x = sqrt (1 / (1 16 t ^ 2)
Back to substitute x = arctan (4T) = 4T and tanx cosx = sqrt (1 / 1 16 t ^ 2)
1 / 4 [x tanx + ln (cos)
1 / 4 [arctan (4T) * 4T + ln [1 / 1 16 t ^ 2] ^ 1 / 2 = 1 / 4 [arctan (1/2ln 4T *) 4T + (1 / 1 16 t ^ 2]
=> T arctan (4T) + (1 / 8) ln (1 / 1 16 t ^ 2) + c
1)
in
. ∫ (lnx) ^ 2 (x ^ 4) dx, before he
try this page:
http://rbmix.com/problem/ix/ix.php
2) to
. ∫ (arctan4t) 1 TD
try this link
http://rbmix.com/problem/in/in.php
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