Can any one solve this integral please?

Question

integration( (24*x^3)/( (4+x^2)^(5/2) ) for x from (2*sqare root of 3) to infinity ) with the subsitution of x = tan u

Answer 1

∫ {24x^3 / [(4 + x^2)^(5/2)]} dx =

as you suggest, let:

x = 2 tan u →

dx = 2 sec^2 u du

thus, substituting, you get:

∫ {24(2 tan u)^3 / [4 + (2 tan u)^2]^(5/2)} 2 sec^2 u du =

∫ {24(8 tan^3 u)  / [4 + (4 tan^2 u) ]^(5/2)} 2 sec^2 u du =

∫ {24(8)(2) tan^3 u  / [4 + (4 tan^2 u) ]^(5/2)}  sec^2 u du =

factor out 4 as:

∫ {24(8)(2) tan^3 u  / [4(1 + tan^2 u)]^(5/2)}  sec^2 u du =

∫ { 24(8)(2) tan^3 u  /{[4^(5/2)](1 + tan^2 u)^(5/2)} } sec^2 u du =

∫ { 24(8)(2) tan^3 u  /[32 (1 + tan^2 u)^(5/2)] } sec^2 u du =

∫ { 12 tan^3 u  /[(1 + tan^2 u)^(5/2)] } sec^2 u du =

12 ∫ { tan^3 u  /[(1 + tan^2 u)^(5/2)] } sec^2 u du =

replace 1 + tan^2 u with sec^2 u, yielding:

12 ∫ { tan^3 u  /[(sec^2 u)^(5/2)] } sec^2 u du =

12 ∫ { tan^3 u  /[(sec^5 u) } sec^2 u du =

then, sec^2 u canceling out, you get:

12 ∫ ( tan^3 u /sec^3 u)  du =

12 ∫ ( tan u /sec u)^3  du =

expressing the integrand in terms of sinx, cosx:

12 ∫ [(sin u/cos u)/ (1 /cos u)]^3  du =

12 ∫ [(sin u/cos u) cos u]^3  du =

cos u canceling out,

12 ∫ sin^3 u du =

rewrite it as:

12 ∫ (sin^2 u) (sin u) du =

replace sin^2 u with (1 - cos^2 u):

12 ∫ (1 - cos^2 u) (sin u) du =

expand and split it into:

12 ∫ (sin u - cos^2 u sin u) du =

12 ∫ sin u du - 12 ∫ cos^2 u sin u du =

as you know, sin u du = - d(cos u), thus:

12 ∫ - d(cos u) - 12 ∫ cos^2 u [- d(cos u)] =

-12 ∫ d(cos u) + 12 ∫ cos^2 u d(cos u) =

-12 cos u + 12 [cos^(2+1) u]/(2+1) + C =

-12 cos u + (12/3) cos^3 u + C =

-12 cos u + 4 cos^3 u + C

summing up,

x = 2 tan u

thus

tan u = (x/2) →

therefore, rewriting cos u in terms of  tan u (i.e. x/2), you get:

sec^2 u = 1 + tan^2 u →

sec u = √(1 + tan^2) →

cos u = 1/√(1 + tan^2) = 1/√[1 + (x/2)^2] = 1/√{1 + [(x^2)/4} = 1/√[(4 + x^2)/4] →

cos u = 2/√(4 + x^2)

thus, substituting back, you get:

-12 cos u + 4 cos^3 u + C  = -12 [2/√(4 + x^2)]  + 4 [2/√(4 + x^2)]^3 + C =

-12 [2/√(4 + x^2)]  + 4 [8/√(4 + x^2)^3] + C =

[- 24/√(4 + x^2)]  + [32 /√(4 + x^2)^3]  + C  that is the antiderivative

then, evaluating your improper integral, you get:

lim {[- 24/√(4 +x^2)] + [32/√(4 +x^2)^3]} - { {- 24/√[4 +(2√3)^2]} + {32/√[4 +(2√3)^2]^3} } =

x → ∞

0 + 0 - { [- 24/√(4 +12)] + [32/√(4 + 12)^3] } =

(24/√16) - [32/√(16)^3]  =

(24/4) - (32/64) = 6 - (1/2) = 11/2

thus your integral converges to (11/2)

(I've just checked this: it's correct)

I hope it helps...

Bye!

Answer 2

Though it can be done by substitution x = 2tanu, it is better to use

4 + x² = u => 2x dx = du

Also, x = 2√3 => u = 4 and x → ∞ => u → ∞

=> ∫(24*x³ )/(4+x^2)^(5/2) dx

= 12 ∫(u -4) / u^(5/2) du

= 12 [u^(-3/2) - 4u^(-5/2)] du

= 12 [ -2u^(-1/2) - (8/3)u^(-3/2) ] + c

Plugging the limits, u = 4 to ∞,

= 12[0 + 1/2 + 1]

= 18.

Answer 3

No, if u = 4 + x², then u(2√3) = 16.

Let x = 2tanu ==> dx = 2sec²u du

u(π/2) -> ∞; u(π/3) = 2√3

...π/2

∫    (24)(8)tan³u)(2)sec²u/(32sec^5u)du

...π/3

...π/2

=∫    (384tan³u)/(32sec³u)du

...π/3

.......π/2

= 12∫    (sin³u)(cos³)/(cos³u) du

......π/3

.......π/2

= 12∫    sin³u du

......π/3

..................................π/2

=12[(-sin²ucosu)/2 -(2/3)cosu]  = 25/4

..................................π/3