How come I cant use the table of integrals

Question

integrals of --> dx/(9x-16x^2)^1/2

I tried to solve it using this dx/(a^2-x^2)^1/2 = arcsin(x/a)+C from the table of integrals but Im not getting the right answer. Why cant I use it or am I not using it correctly?

Answer 1

Because a^2 is a perfect square constant. however, your function says 9x - 16x^2, which has a LINEAR first term, not a constant one. So the formula you are trying to use does not apply.

Rewrites the denominator:

9x-16x^2 = -16x^2 + 9x = -16(x^2 - (9/16)x)

= -16( x^2 - (9/16)x + 81/1024) + 81/64 <---- completes the square

= 81/64 - 16(x^2 - (9/16)x + 81/1024))

= 81/64 - (16x^2 - 9x + 81/64) <--- distributes the 16 but NOT the negative

= 81/64 - (4x - 9/8)^2

= 81/64 - (32x - 9)^2/64

= [81 - (32x-9)^2 ]/64

= [ 1 - (32x-9)^2/81] / 64

= [ 1 - ( (32x-9)/9)^2]/64

taking the square root of this whole thing:

sqrt [ 1 - ( (32x-9)/9)^2]]/8

however, this expression is in the denominator, so the 8 rises to the numerator.

Substiution U = (32x-9)/9 gives the standard form for the integral arcsin

sqrt [ 1 - U^2]

however, dU = 32/9 dx so dx = 9 * dU/ 32

The 32 cancels the 8 on top, so the 4 stays in the denominator.

The final integral is

(1/4) integral ( dU / (1-U^2)^1/2)

which integrates to (1/4) arcsin(U) = (1/4) arcsin( (32x-9)/9)