calculus help?

Question

1. ∫ x √(1-x^4) dx

Answer 1

1) ∫ x√(1 - x^4) dx

First, express x^4 as (x^2)^2.

∫ x√(1 - (x^2)^2) dx

I'm going to move the x next to the dx to make the next step more obvious.

∫ √(1 - (x^2)^2) x dx

Here's where we use substitution.

Let u = x^2. Then
du = 2x dx, and if we multiply both sides by (1/2), we get

(1/2)du = x dx {Note: x dx is the tail end of our integral, so (1/2)du is going to be the tail end of our new integral after the substitution.

∫( √(1 - u^2) (1/2) du )

Pull the (1/2) out of the integral ;after all, it's a constant.

(1/2) ∫( √(1 - u^2) du )

At this point, we have no choice but to use trigonometric substitution.

Let u = sin(t). Then
du = cos(t) dt

(1/2) ∫( √(1 - sin^2(t)) cos(t) dt )

Use the trig identity to change the inside of the square root to cos^2(t).

(1/2) ∫( √(cos^2(t)) cos(t) dt )

The square root of a square is just itself.

(1/2) ∫( cos(t) cos(t) dt )

Merge these into one.

(1/2) ∫( cos^2(t) dt )

To solve this, we need to use the half angle identity. Note that
cos^2(t) = (1/2) (1 + cos(2t))

(1/2) ∫( (1/2) (1 + cos(2t)) dt )

Pull out the (1/2)

(1/4) ∫ [ (1 + cos(2t)) dt ]

This is now a more integrable form. Remember that the integral of cos(kt) is (1/k)sin(kt).

(1/4) [t + (1/2)sin(2t)] + C

Distributing the (1/4), and we get

(1/4)t + (1/8)sin(2t) + C

Using the double angle identity sin(2t) = 2sin(t)cos(t), we get

(1/4)t + (1/8)2sin(t)cos(t) + C, which reduces to

(1/4)t + (1/4)sin(t)cos(t) + C

Now, we have to make everything in terms of x. Recall that
u = sin(t). This means

sin(t) = u = u/1 = opp/hyp (by the SOHCAHTOA RULES). This means

opp = u
hyp = 1, and by Pythagoras,
adj = sqrt(hyp^2 - opp^2) = sqrt(1 - u^2)

cos(t) = adj/hyp = sqrt(1 - u^2)/1 = sqrt(1 - u^2)
Also, u = sin(t) implies t = arcsin(u), so

(1/4)t + (1/4)sin(t)cos(t) + C becomes

(1/4)arcsin(u) + (1/4)[u][sqrt(1 - u^2)] + C

But u = x^2, so our final answer is

(1/4)arcsin(x^2) + (1/4)x^2 sqrt(1 - x^4) + C

Answer 2

∫ x √(1-x^4) dx
= ∫ (1/2) √(1-x^4) dx^2
= (1/4)[x^2√(1-x^4) + arcsin(x^2)]
--------
Reason:
∫ √(1-x^2) dx
= x √(1-x^2) + ∫(x^2-1+1)/ √(1-x^2) dx, integration by parts
= x √(1-x^2) - ∫ √(1-x^2) dx + arcsin(x^2)

∫ √(1-x^2) dx, collect like terms and solve for ∫ √(1-x^2) dx
= (1/2)[x √(1-x^2) + arcsin(x^2)]

Answer 3

Integrate ∫ x √(1 - x^4) dx.

∫ x √(1 - x^4) dx
Let
sinθ = x²
cosθ dθ = 2x dx
(1/2)cosθ dθ = x dx

∫ x √(1 - x^4) dx
= (1/2)∫ cosθ √(1 - sin²θ) dθ
= (1/2)∫ cosθ √(cos²θ) dθ
= (1/2)∫ cos²θ dθ
= (1/4)∫ [(2cos²θ -1) + 1] dθ
= (1/4)∫ [(cos 2θ) + 1] dθ
= (1/8)(sin 2θ) + θ/4 + C
= (1/4)(sinθ)(cosθ) + θ/4 + C

sinθ = x²
θ = arcsin(x²)
cosθ = √(1 - sin²θ) = √(1 - x^4)

= (1/4)(sinθ)(cosθ) + θ/4 + C
= (1/4) { x²√(1 - x^4) + arcsin(x²) } + C

Answer 4

( (√(1 - x^4)) / ( x^2( ( 4*(1-x^4) / x^4 ) + 4 ) ) - ( (atan( (√(1 - x^4)) / x^2) ) / 4 )

simplified : ( (x^2√(1 - x^4)) - (atan( (√(1 - x^4)) / x^2) ) ) / 4