∫ 1 / ((x^2)*(x^2-25)^(1/2)) dx and ∫ 1 / (16-x^2)^(5/2)?

Question

∫ 1 / ((x^2)*(x^2-25)^(1/2)) dx and ∫ 1 / (16-x^2)^(5/2)

Answer 1

(i)
∫ 1 / [x² · sqrt(x²-25)] dx
Integrals containing functions of sqrt(x² - a²) can be transferred to elementary integrals by the substitution
x = a · cosh(t)

So substitute
x = 5 · cosh(t)
dx = 5 · sinh(t) dt = 5 ·sqrt(cosh²(t) - 1)

∫ 1 / [x² · sqrt(x²-25)] dx
= ∫ [5 ·sqrt(cosh²(t) - 1)] / [25 · cosh²(t)· sqrt(25 · cosh²(t)²-25)] dt
= ∫ 1 / [25 · cosh²(t) dt
= tanh(t) / 25 + c
= sinh(t) / [25 · cosh(t)] +c
= sqrt(cosh²(t) - 1) / [25 · cosh(t)] +c
(back substitution with cosh(t)] = x/5)
= sqrt(x²/25 - 1) / [25 · x/5] +c
= sqrt(x² - 25) / [25 · x] +c

(ii)
∫ 1 / (16-x²)^(5/2) dx
Integrals containing functions of sqrt(a² - x²) can be transferred to elementary integrals by the substitution
x = a · sin(t)

So substitute
x = 4 · sin(t)
dx = 4 · cos(t) dt = 4 ·sqrt(1 - sin²(t)) dt

∫ 1 / (16-x²)^(5/2)
= ∫ 1 / [sqrt(16-x²]^5 dx
= ∫ 4·sqrt(1 - sin²(t)) / [sqrt(16 - 16·sin²(t))]^5 dx
= ∫ 1 / [4 ·sqrt(1 - sin²(t))]^4 dt
= 1/256 · ∫ 1 / cos^4(t) dt
= 1/256 · (1/3 · sin(t) / cos³(t) + 2/3 · sin(t) / cos(t) ) + c
= 1/768 · (sin(t) / [sqrt(1- sin²(t)]³ + 2 sin(t) / [sqrt(1- sin²(t)] ) + c
Back substitution sin(t) = x/4
= 1/768 · (x/4 / [sqrt(1 - x²/16)]³ + 2· x/4 / [sqrt(1 - x²/16)] ) + c
= 1/768 · (16 ·x / [sqrt(16 - x²)]³ + 2 · x / [sqrt(16 - x²)] ) + c
= 1/384 · (8 ·x / [sqrt(16 - x²)]³ + x / [sqrt(16 - x²/)] ) + c

Answer 2

hmmm...probably...however the single that fits best for the wizard or the single he's best suitable for stands out as the single chosen. it relatively is not likely nonetheless that the point of compatibility between the two wands would be realtively extreme! One is definite to be extra nicely matched. additionally, each and each wand has a center, and a definite lentgh. The length of a wand relies upon on the dimensions of the wizards arm, and the middle ultimately relies upon on what form of magic the wizard prefers. If a wizard is extra apt for Transfiuration , as an occasion, then that individual would be extra effective to a definite center. and additionally of direction if the wand is gained, then that wand provides you with its allegiance...making you an proprietor of two wands...very like Harry and Dumbledore! So i think of its accessible, yet fairly no longer likely! And it very hardly occurs (i think of)

Answer 3

♥ leaving us in voting? I don’t like it man!

Answer 4

.....four.