Integration Help?

Question

Integrate this......

3 / sqr (9 - 4x^2) dx

Show the answer is

3/2sineinverse ( 2x / 3) = C

Can anyone help?

Answer 1

start with integrating 1/(sqrt(1-x^2))

the substitution is x = sin(u)
dx= cos(u) du

the denominator becomes sqrt(1- sin^2(u)) = sqrt(cos^2(u))=cos(u)

so the integral becomes cos(u) du/cos(u) which is the integral of 1du = u which is arcsine(x)

in your example you have a 3 in the numerator. take it out and multiply it at the end.

instead of sqrt(1-x^2) in the denominator, you have (9-4x^2) in the denominator.
see if you can find the substitution similar to the above to reduce the problem.

Answer 2

∫ [3 / sqrt(9 - 4x^2)] dx

Let x = (3/2)sin(t)
Then dx = (3/2)cos(t) dt, so we have

∫ [ 3 / (sqrt(9 - 4[(9/4)sin^2(t)]) ] [ (3/2)cos(t) ] dt

Reducing this, we get

∫ [ 3 / (sqrt(9 - 9sin^2(t)) ] [ (3/2)cos(t) ] dt

Factoring the 9 within the square root lets us take the 9 out as a 3.

∫ [ 3 / 3(sqrt(1 - sin^2(t)) ] [ (3/2)cos(t) ] dt

which we can further reduce. We can also change 1 - sin^2(t) into cos^2(t). Notice the 3 on the top and bottom cancel, too.

∫ [1 / sqrt(cos^2(t)] [(3/2) cos(t)] dt

Taking the square root of a square will give us just its singular form.

∫ [1 / cos(t)] [(3/2) cos(t)] dt

Now, we can pull the constant (3/2) out of the integral.

(3/2) ∫ [1 / cos(t)] [cos(t)] dt

Note that cos(t) and 1/cos(t) cancel each other out, to 1.

(3/2) ∫ (1) dt

Which is easily integrable now, as

(3/2)t + C

And now, since we let x = (3/2) sin(t), it follows that

(2/3)x = sin(t), and taking the arcsin of both sides to isolate the t, we have
arcsin ( (2/3)x ) = t

Therefore, substituting that value in for t gives

(3/2) arcsin( (2/3)x ) + C