sqrt (16 - x^2) all over x^2 dx?

Question

Trigonometric Integration - Step by step

Answer 1

well Robert i just give you the correct answer, b coz they explained the correct way;
I just dont waste your time more;

∫√(16 - x^2) /x^2 dx =
-(√(16 - x^2) / x) - ArcSin(x/4) +c

Good Luck Rob..

Answer 2

Realize that
sqrt(16 - x^2) = 4 * sqrt(1 - (x/4)^2)

Compare this with
cos t = sqrt(1 - [sin t]^2),

So if we write x = 4 sin t, we get

[4 cos t / (4 sin t)^2] d(4 sin t)
=
[cos^2 t / sin^2 t] dt = cot^2 t dt

Since d(cot t)/dt = -cot^2 t - 1, we have
cot^2 t dt = -1 - d(cot t)/dt = -d(t - cot t)/dt

so the integral becomes
- t - cot t + C = - 1/4 asin x - sqrt(16 - x^2)/x + C

Answer 3

I know this by form:

f(x) = integral of (sqrt(16-x^2)/x^2

f(x) = -sqrt(16 - x^2)/x - asin(x/4) + C

asin = inverse sin.

To Check:

d(asin(x/4))/dx = 1/sqrt(16-x^2)

d(f(x))/dx = (1/2)(1/x)(2x)(1/sqrt(16-x^2)) + sqrt(16-x^2)/x^2 - 1/sqrt(16-x^2)

The first term simplifies to:
1/sqrt(16-x^2) which cancels the d(asin(x/4))/dx term and you are left with: sqrt(16-x^2)/x^2