Integral of these 2 problems?

Question

1. integral of dx/x sqrt(1-x^4)

i tried using teh substitution u=x^2 to make it look like arcsine xut the x on teh bottom is a problem. the arcsecant methid isnt working either.

2.integral of ln(arccos x)dx/(arccosx)(sqrt (1-x^2))

i used the substitution u=arccos x and i ended up with the negative integral of ln u/u du

the problem is that i have to use yet another substitution. i end up with the negative integral of ydy(y being the term i used in teh second substitution)

pls help

Answer 1

Why can't people learn to use parenthesis?

I guess this is:

dx/[x*sqrt(1 - x^4)]

u = x^2

du = 2xdx and dx = du/2x

dx/[x*SQRT(1 - x^4)] = du/[(2u)SQRT(1 - u^2)]

Now use u = sin(w) and du = cos(w)dw

Also 1 - u^2 = 1 - sin^2(w) = cos^2(w)

du/[(2u)SQRT(1 - u^2)] = cos(w)dw / [2sin(w)cos(w)]

du/[(2u)SQRT(1 - u^2)] = (1/2)[1/sin(w)]dw

Integral = (1/2)ln[tan(w/2)] or (1/2)ln[csc(w) - cot(w)]

u = sin(w)

cot(w) = SQRT(1 - u^2)/u

csc(w) = 1/sin(w) = 1/u

Integral = (1/2)ln[1/u - SQRT(1 - u^2)/u]

Integral = (1/2)ln[1/x^2 - SQRT(1 - x^4)/x^2]

Integral = (1/(2)ln{(1/x^2)[1 - SQRT(1 - x^4)]}

Check: Differentiate the answer and see if we get the original equation.

Let g = the stuff inside the ln for ease in writing this

d[(1/2)ln(g)] -> (1/2)(1/g)dg

g = [1/x^2 - SQRT(1 - x^4)/x^2] =  (1/x^2)[1  - SQRT(1 - x^4)]

dg = {(-2/x^3)[1 - SQRT(1 - x^4)] - (1/x^2)[-(2x^3/SQRT(1 - x^4)]} dx

dg/g = [(-2/x) + 2x^3/{[1  - SQRT(1 - x^4)][SQRT(1 - x^4)] }]dx

Put the first term over the same denominator as the second:

(-2/x){[1 - SQRT(1 - x^4)][SQRT(1 - x^4)]} + 2x^3

(-2/x){[1 - SQRT(1 - x^4)][SQRT(1 - x^4)] - x^4}

(-2/x){SQRT(1 - x^4) - (1 - x^4) - x^4}

(2/x){1 - SQRT(1 - x^4)}

Bring back the denominator and stuff:

(2/x){1 - SQRT(1 - x^4)}dx / {[1  - SQRT(1 - x^4)][SQRT(1 - x^4)] }

(1/2)(1/g)dg = dx/[x*SQRT(1 - x^4)] which is the original equation so the solution is good

2. there is nothing wrong with what you did since I ended up with the same thing. You have a ydy which integrates to y^2/2 and y is ln(u) and so on. perfectly OK.

u = arccos(x)

x = cos(u) so sin(u) = SQRT(1 - x^2)

dx = [-sin(u)]du

ln(u)[-sin(u)du] / [u*sin(u)] = -ln(u)du/u

But d[ln(u)] = (1/u)du so:

w = ln(u) and dw = (1/u)du

Integral = -ln(u)du/u = -wdw

Integrated this becomes -w^2/2 = -[ln(u)]^2/2  = (-1/2){ln[arccos(x)]}^2

Check: differentiate this to see if we get what we started with.

(-1/2)(2)ln[arccos(x)][1/arccos(x)][-1... - x^2)]dx

ln[arccos(x)][1/arccos(x)]dx[1/SQRT(1 - x^2)]

And this is the original equation so again the solution is good

I used the tables from the listed site.

Note. there is nothing wrong with using more than one substitution.