Integrate 1/(x*sqrt(a-bx^2))

Question

Hi there,

I've been toiling for hours trying to answer this one:

Integrate 1/(x*sqrt(a - bx^2))

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is 1/sqrt(a)*log((sqrt(a)-sqrt(a-bx^2))/(x*sqrt(a)))

I've attempted the question by letting x=sqrt(a/b)*sin(u), which I *think* is on the right track as I end up with the 1/sqrt(a) term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew

Answer 1

Let U = sqrt(a - bx^2) = (a - bx^2)^(1/2)

**Note that as a result, U^2 = a-bx^2 and then a-U^2 = bx^2

then dU = (1/2)(a - bx^2)^(-1/2) (2bx) dx

= (a-bx^2)^(-1/2) (bx) dx

= (bx)*dx/ sqrt(a - bx^2)

So then dx = dU *sqrt(a-bx^2)/(bx)

The integral in question becomes:

dU / bx^2 =

dU / (a-U^2) <--- **from the note specified above

This can be integrated via partial fraction decomposition (pfd):

I(U) = dU / (k+ U)(k - U)) where k = sqrt(a)

PFD is :

I(U)= B/(k+U) + C/(k-U) where k,C, and B are fixed # constants

but k is known and given

Multiplies both sides by LCD (k+U)(k-U)

1 = B(k-U) + C(k+U)

1 = k(B+C) + U(C-B) <-- remember , var is U

then C-b= 0

(B+C)k = 1

The first equation says b=c

the second equation becomes: 2bk=1

b = 1/(2k)

the pfd is:

(1/2k) [ 1/(k+U) + 1/(k-U)]

Integrating linearly:

(1/2k) ln[k+u] + ln[k-U] where U is the square root function

substituted above

integration complete

Answer 2

The u-substitution for this integral is not obvious. Sorry I do not have time to invest in a full

solution. I did a little digging and came up with this:

Let u = sin^-1((√b/√a)x) --> cos(u) du = (√b/√a) dx

This leads to (1/√a) ∫ csc(u) du which in turn has a non-obvious method.

See http://math2.org/math/integrals/more/csc.htm