HARD INTEGRATION: What is the integral (x^2)/(sqrt(9-25(x^2))dx with bounds 0.6 and 0?

Question

No stupid remarks or answers please. This is the 4th time I put this question up because people don't take the question seriously. Any Serious Mathematicians are welcome!! Also, if you answer first and have the right answer and method I will automatically make your response the best answer (giving you more points).

Answer 1

This is a straightforward trig substitution.

Let sinθ = 5/3x (so that x² = 9/25 sin²x); then

3/5 cosθ dθ = dx, and

3cosθ = √(9-25x²).



This means that, when all the algebra is out of the way, your integral is the same as

∫9/125 sin²θ dθ. (Assuming my arithmetic was correct, anyway.)



We can also change limits; when x=0, θ=0; when x=0.6, sinθ = 1, so θ=π/2.



Note that

∫sin²θ dθ = θ/2 - 1/2 sinθ cosθ + C,

so (by my count) this gives a final value of

9/125 (π/4) = 9π / 500.

§



** REMARK: **



While this is technically an improper integral--the denominator is 0 at the right endpoint in the original integral--the limit of the integral exists, and so the value stands. In fact, the trig-substituted version is well-defined over the whole interval!

Answer 2

Let 5x = 3sin(u) so

dx = 3/5 cos(u) du. Then the integral becomes

int_{0}^{pi/2} 3/5*cos(u)*9/25*sin^2(u)/

(sqrt(9 - 9sin^2(u)) du

where _{ } and ^{ } means lower and upper limit of integration. Note when x =0, sin(u) =0 so u =0 and when x =0.6 = 3/5, sin(u) = 1, so u = pi/2.

The integrand simplifies considerably since 1 - sin^2(u) = cos^2(u) and the integral becomes

3/5*9/25*1/3* int_{0}^{pi/2} sin^2(u) du

= 9/125* int_{0}^{pi/2} 1/2[1 - cos(2u)] du

= 9/125*1/2* [u - 1/2sin(2u)]_{0}^{pi/2}

= 9/250*[pi/2 - 1/2sin(pi) - (0 - 1/2sin(0))]

= 9pi/500.

Answer 3

∫[x^2 / sqrt(9 - 25x^2)]dx



let x = (3/5)siny : siny=5x/3: cosy=(1/3)sqrt(9- 25x^2)y= sin^-1(5x/3)



dx = (3/5)cosy dy



now the integral becomes



∫(3/5)^2 sin^2(y) (3/5) cosy dy / sqrt(9 - 9sin^2(y)



=>27/125∫sin^2(y)cosy dy / 3(cosy)



=>9/125∫sin^2(y) dy



=>9/125∫[(1- cos(2y)/2] dy



=>9/250∫dy - 9/250∫cos(2y) dy



=> (9/250)y - (9/250)(1/2)(sin(2y)



=>(9/250)y - (9/250) siny cosy



substituting back siny=5x/3: cosy=(1/3)sqrt(9- 25x^2)y= sin^-1(5x/3)



=>(9/250)sin^-1(5x/3) -(1/50)x *sqrt(9-25x^2)



applying limits 0 to 0.6



(9/250)[sin^-1(1) - sin^-1(0) - (1/50)[0.6sqrt(9 - 9) - 0]



(9/250)[pi/2]



=>9 pi/500