Some calc questions

• May 13th 2006, 12:55 PM
nirva
Some calc questions
I'm having problems with some definite intergral questions

Int2-1 (Ln(x)/x^2) dx
I got 1 as a result from that, but it's different from the answer, I'm wondering about how to do it?

Also
Int cos(Ln(x)) dx

Can anyone also tell me what the derivative of cos^-1, tan^-1, sin^-1.. etc
• May 13th 2006, 04:14 PM
ticbol
Quote:

Originally Posted by nirva
I'm having problems with some definite intergral questions

Int2-1 (Ln(x)/x^2) dx
I got 1 as a result from that, but it's different from the answer, I'm wondering about how to do it?

Also
Int cos(Ln(x)) dx

Can anyone also tell me what the derivative of cos^-1, tan^-1, sin^-1.. etc

1) INT.(2->1)[lnX / X^2]dX --------------(1)
By parts.
INT.[u]dv = uv -INT.[v]du ------(i)

Let u = lnX
so, du = (1/X)dX

And dv = dX / X^2
So, v = -1/X

Then, for the indefinite integration of (1),
INT.[lnX / X^2]dX
= (lnX)(-1/X) -INT.[-1/X](dX /X)
= -(lnX)/X +INT.[1/X^2]dx
= -(lnX)/X +(-1/X) +C
= -(LnX)/X -1/X +C
And so for the definite integration of (1),
= -[(lnX)/X +1/X](2->1)
= -[ln(1)/1 +1/1] +[ln(2)/2 +1/2]
= -1 +(1/2)ln(2) +1/2

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2) INT.[cos(lnX)]dx -----(2)

Whoa, this could be long, so let me skip some steps.

Let U = lnX
So, X = e^U

and, dU = dX /X
dX = X*dU = e^U dU

Then, substitute those into (2),
INT.[cos(lnX)]dX
= INT.[cosU](e^U dU) ------(2a)

Integrate (2a) by parts:
INT.[r]ds = r*s -INT.[s]dr
r = e^U; so, dr = e^U dU
ds = cosU dU; so, s = sinU
Then, back to (2a),
= e^U *sinU -INT.[sinU](e^U dU)
= e^U *sinU -INT.[e^U](sinU dU) -------(2b)

Integrate the last part of (2b) by parts again,
INT.[m]dn = m*n -INT.[n]dm
m = e^U; so, dm = e^U dU
dn = sinU dU; so, n = -cosU
Then, back to (2b),
= e^U *sinU -{e^U *(-cosU) -INT.[-cosU](e^U dU)}
= e^U *sinU +e^U *cosU -INT.[cosU](e^U dU) -------(2c)

The last part of (2c) is the same as (2a),
And since (2a) is (2) also, then,
= e^U *sinU +e^U *cosU -INT.[cos(lnX)]dX

Transpose the last part to the unseen LHS,
2(INT.[cos(lnX)]dX) = e^U *sinU +e^U *cosU
So, divide both sides by 2,
INT.[cos(lnX)]dX = (1/2)[e^U *sinU +e^U *cosU]
Substituting back away from U,
INT.[cos(lnX)]dX = (1/2)[X*sin(lnX) +X*cos(lnX)]

Therefore,
INT.[cos(lnX)]dX = (X/2)[sin(lnX) +cos(lnX)] +C. --------answer

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For general derivatives or integrals, go to this website:
http://www.sosmath.com/tables/tables.html
(Sorry, I do not know how to link.)

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