Tiny integral with huuuuuuge integration?

∫ (9 + 4/√x )^3/2 dx

I like the suggestion of subbing x = cot^4

how abotu this

sub x = 16cot^4(u)/81

(so √x = 4cot²(u)/9, and 4/√x = 9tan²(u))

dx = -64cot²(u)csc²(u) du / 81

∫ (9 + 4/√x )^3/2 dx

= ∫ (9 + 9tan²(u))^3/2 (-64cot²(u)csc²(u) du / 81)

= -64/3 * ∫ (1 + tan²(u))^3/2 (cot²(u)csc²(u) du)

= -64/3 * ∫ (sec²(u))^3/2 cot²(u)csc²(u) du

= -64/3 * ∫ sec³(u)cot²(u)csc²(u) du

= -64/3 * ∫ cot(u)csc³(u)du

= 64/3 * ∫ csc²(u) [-cot(u)csc(u)du]

sub v = csc(u)

dv = -csc(u)cot(u)du

= 64/3 * ∫ v² dv

= 64v³/9 + C

= 64csc³(u)/9 + C

= 64csc³(u)/9 + C

x = 16cot^4(u)/81

cot^4(u) = 81x/16

cot(u) = 3x^(1/4)/2 (=A/O)

csc(u) = H/O

= √(9√x + 4) / 2

= 64√(9√x + 4)³/72 + C

Sorry, I really rushed this due to lack of time and apparently this answer is incorrect (according to online integrators) but I don't have time to fix it now and I leave the city for a week tomorrow So I can't come back and check it (although if I find a computer to use I will try)

I do believe in this method though, might be worth someone revisiting.

I'll make a couple of substitutions, the first involving a power function and the second involving a hyperbolic function. Will skip over many details since you guys are obviously calculus experts. :)

First, let x = (16/81)t^4 then the integral becomes ∫ (t^2+1)^(3/2) (64/3) dt. Let's temporarily ignore the multiplicative constant in the following work for convenience.

Next, let t = sinh(p), dt = cosh(p) dp. Note that cosh²(p) = 1 + sinh²(p). So the integral now becomes ∫ (cosh^4)(p) dp. Note that by definition cosh(p) = (e^p + e^(-p))/2. So we can do a binomial expansion of (cosh^4)(p) to get [(e^(4p) + e^(-4p)) + 4(e^(2p) + e^(-2p)) + 6]/16. Each of the curve-bracketed terms can then be converted back to cosh'es i.e. [2cosh(4p) + 8cosh(2p) + 6]/16. This can now be easily integrated as sinh(4p)/32 + sinh(2p)/4 + 3p/8.

So now we just have to reverse both substitutions. Let inv_sinh be the inverse hyperbolic sine (it is a logarithm). First the hyperbolic sub: sinh(4 inv_sinh(t))/32 + sinh(2 inv_sinh(t))/4 + 3 inv_sinh(t)/8. Note that it is possible to simplify the sinh-inv_sinh terms but I won't bother here. Next the power sub: sinh(4 inv_sinh((3/2)x^(1/4)))/32 + sinh(2 inv_sinh((3/2)x^(1/4)))/4 + 3 inv_sinh((3/2)x^(1/4))/8

So the final answer should be (64/3)[sinh(4 inv_sinh((3/2)x^(1/4)))/32 + sinh(2 inv_sinh((3/2)x^(1/4)))/4 + 3 inv_sinh((3/2)x^(1/4))/8] + C

Hopefully there are no numerical errors but let me know if there are.

hey Tabla,

how about a Trigonometric/Hyperbolic substitution?

ie.

for simplicity I will just consider

int ( ( 1 + 1/sqrt(x) ) ^ (3/2) dx )

let x = cot^4(s) -->dx = -4cot^3(s)cosec^2(s)

Thus,

int( ( 1 + tan^2(s))^(3/2) * -4cot^3(s)cosec^2(s) ds)

= -4*int( sec^3(s) * cot^3(s)cosec^2(s) ds )

= -4*int(cosec^5(s) ds)

which is a pain in the **** with Integration by parts, but achievable

What are your thoughts?

David

How does the chain rule work on on that one?