Help solving integral...?

5 - x^2 = u

-2x * dx = du

x^5 * (5 - x^2)^(1/2) * dx =>

x^2 * x^2 * (5 - x^2)^(1/2) * x * dx =>

(-1/2) * x^2 * x^2 * (5 - x^2)^(1/2) * (-2 * x * dx) =>

(-1/2) * (5 - u) * (5 - u) * u^(1/2) * du =>

(-1/2) * (25 - 10u + u^2) * u^(1/2) * du =>

(-1/2) * (25 * u^(1/2) - 10 * u^(3/2) + u^(5/2)) * du

Now it's just a matter of using the power rule to integrate:

(-1/2) * (25 * (2/3) * u^(3/2) - 10 * (2/5) * u^(5/2) + (2/7) * u^(7/2)) + C

(-1/2) * 2 * u^(3/2) * ((25/3) - (10/5) * u + (1/7) * u^2) + C

-1 * u^(3/2) * ((25/3) - 2u + (1/7) * u^2) + C

-u^(3/2) * (1/3) * (1/7) * (25 * 7 - 2 * 3 * 7 * u + 3 * u^2) + C

-u^(3/2) * (1/21) * (175 - 42u + 3u^2) + C

(-1/21) * (3u^2 - 42u + 175) * u^(3/2) + C

(-1/21) * (3 * (5 - x^2)^2 - 42 * (5 - x^2) + 175) * (5 - x^2)^(3/2) + C

(-1/21) * (3 * (25 - 10x^2 + x^4) - 210 + 42x^2 + 175) * (5 - x^2)^(3/2) + C

(-1/21) * (75 - 30x^2 + 3x^4 - 210 + 42x^2 + 175) * (5 - x^2)^(3/2) + C

(-1/21) * (3x^4 + 12x^2 + 40) * (5 - x^2)^(3/2) + C

That's about as nice as it's going to get.

Use u = 5-x^2, then du = -2xdx.

Note that in part of the solution, we need to express x^4 as a function of u = 5 - x^2.

integral x^5*(5-x^2)^(1/2) dx

= integral [x^5/(-2x)] * (5-x^2)^(1/2) * (-2x) dx, from multiplying and dividing by -2x

= integral (-1/2)x^4 * (5-x^2)^(1/2) * (-2x) dx

= integral (-1/2)(x^2)^2 * (5-x^2)^(1/2) * (-2x) dx

= integral (-1/2)[5 - (5 - x^2)]^2 * (5-x^2)^(1/2) * (-2x) dx

= integral (-1/2)(5 - u)^2 * u^(1/2) du

= integral (-1/2)(u^2 - 10u + 25) * u^(1/2) du

= integral (-1/2)(u^(5/2) - 10u^(3/2) + 25u^(1/2)) du

= (-1/2)((2/7)u^(7/2) - 4u^(5/2) + (50/3)u^(3/2)) + C

= (-1/7)(5-x^2)^(7/2) + 2(5-x^2)^(5/2) - (25/3)(5-x^2)^(3/2) + C

Lord bless you today!

∫ x^5 √(5 - x^2) dx

= ∫ x^4 √(5 - x^2) x dx

let 5 - x^2 = t^2

x^2 = 5 - t^2

2x dx = -2t dt

x dx = -t dt

now the integral changes to

∫ ( 5 - t^2)^2 √t^2 (- t dt )

= -∫ (25 + t^4 - 10t^2) t^2 dt

= -∫ (25t^2 + t^6 - 10t^4) dt

= -(25/3)t^3 - (1/7)t^7 + 2 t^5 + C

= -t^3 /21 [ 3 t^4 - 42 t^2 + 175 ] + C

back substitute

t^2 = 5 - x^2

= - (1/21)(5 - x^2)^(3/2) [ 3 (5 - x^2)^2 - 42(5 - x^2) + 175 ] + C

= - (1/21)(5 - x^2)^(3/2) [ 75 + 3x^4 - 30x^2 - 210 + 42 x^2) + 175 ] + C

= - (1/21)(5 - x^2)^(3/2) [ 3x^4 + 12x^2 - 35 ] + C

don't comprehend what's up with that y^5 up there, yet right here is going your quintessential... int y^4 from 0 to 2 = y^5/5 2^5/5 - 0.5/5 32/5 - 0 quintessential = 32/5 i would be unable to describe right here the thank you to resolve integrals. there are a number of diverse strategies for many categories of integrals. Yours is the finest, yet explaining why and how would take an prolonged, long term.

a little tricky

let x = 5^1/2 sinw dx = 5^1/2cosdw

sub to get [ 5^1/2 sinw ]^5[5^1/2][cosw][5^1/2] cosw dw A = all of constant terms

A[sin^2w]^2 sinw cos^2wdw

A [1-cos^2w]^2cos^2wsinwdw

u = cosw A[1-u^2]^2 u^2 du rest is east

∫x^5 * √(5 - x^2) dx

u = 5 - x^2

du/-2 = x dx

-1/2*∫(5 - u)^2*√u du

-1/2*∫(25 - 10u + u^2)*√u du

-1/2*∫(25√u - 10u^(3/2) + u^(5/2)) du

-25/3*(5 - x^2)^(3/2) + 2*(5 - x^2)^(5/2) - 1/7*(5 - x^2)^(7/2) + C