# Integration by U substitution pls help?

What's the

1-integration of U sqrt 1-U^2 du and

2-integration of sqrt X^3-3/sqrt X^11 dx

### 3 Answers

- cidyahLv 76 years agoBest Answer
1) ∫ u sqrt(1-u^2) du

Let t=1-u^2

dt = -2udu

u du = (-1/2) dt

∫ u sqrt(1-u^2) du = (-1/2) ∫ sqrt(t) dt = (-1/2) ∫ t^(1/2) dt = (-1/2) t^(1/2+1)/(1/2+1) = (-1/2)(2/3) t^(3/2)

= (-1/3) t^(3/2)

= (-1/3) (1-u^2)^(3/2) + C

2) ∫ sqrt(x^3-3) / sqrt(x^11) dx

= ∫ sqrt(x^3-3) / x^(11/2) dx

Factor out x^3 from the square root

= ∫ x^(3/2) sqrt(1-3/x^3) / x^(11/2) dx

= ∫ sqrt(1-3/x^3) / x^(8/2) dx

= ∫ sqrt(1-3/x^3) / x^4 dx

Let u = 1-3/x^3

u=-3 x^(-3)

du=(-3)(-3) x^(-4)

du = 9/x^4 dx

1/x^4 dx = (1/9) du

∫ sqrt(1-3/x^3) / x^4 dx = (1/9) ∫ sqrt(u) du

= (1/9) ∫ u^(1/2) du

= (1/9) u^(1/2+1)/(1/2+1)

= (1/9)(2/3)u^(3/2)

=(2/27) u^(3/2)

= (2/27) (1-3/x^3)^(3/2) + C

- ComoLv 76 years ago
1.

I = ∫ x (1 - x²)^(1/2) dx

Let u = 1 - x²

du = -2x dx

-du/2 = x dx

I = (-1/2) ∫ u^(1/2) du

I = (-1/3) (2/3) u^(3/2) + C

I = (-2/9) (1 - x²) + C

2.

As given MUST be read as x³ - [ 3/x^(11) ] so will answer accordingly.

I = ∫ x³ - 3 x^(11/2) dx

I = x^4 / 4 - (6/13) x^(13/2) + C

- bskelkarLv 76 years ago
1) 2[(1-u^2)^(3/2)]3 + c. Use appropriate formula.

2) Can't understand sqrt(x^3-3) or sqrt(x^3) - 3......