Question

Question #55487

Answer 1

`arcsin(4x)/512+xsqrt(1-16x^2)*(32x^2-1)/256+C`

Explanation:

To find `int\ x^2sqrt(1-16x^2)\ dx`, first substitute `4x=sin(u)`, or `dx=cos(u)/4\ du`.

This gives `1/64int\ sin^2(u)cos(u)sqrt(1-sin^2(u))\ du`. Using the identity `1-sin^2(u)=cos^2(u)`, this can be further simplified to `1/64int\ sin^2(u)cos^2(u)\ du`.

Now, we will use the identities `sin^2(u)=(1-cos(2u))/2` and `cos^2(u)=(1+cos(2u))/2`.

After simplifying with the product of the sum and difference formula, this becomes `1/256int\ 1-cos^2(2u)\ du`. Use the identity `1-cos^2(u)=sin^2(u)` to get `1/256int\ sin^2(2u)\ du`.

Now, we will use the identity `sin^2(u)=(1-cos(2u))/2` again. After a bit of simplifying, we have `1/512int\ 1-cos(4u)\ du`.

Using a bit of integration, the result is `u/512-sin(4u)/2048+C`.

Use the identities `sin(2u)=2sin(u)cos(u)` and `cos(2u)=cos^2(u)-sin^2(u)` to simplify `sin(4u)=2sin(2u)cos(2u)=2sin(u)cos(u)(cos^2(u)-sin^2(u))=2sin(u)cos^3(u)-2sin^3(u)cos(u)`.

Thus, our result simplifies to `u/512-(sin(u)cos^3(u)-sin^3(u)cos(u))/1024+C`.

Now, we know that `4x=sin(u)`, or `u=arcsin(4x)`. We substitute this back in, while remembering that `sin(arcsin(4x))=4x` and `cos(arcsin(4x))=sqrt(1-sin^2(arcsin(4x)))=sqrt(1-16x^2)`.

Thus gives `arcsin(4x)/512-(4x(1-16x^2)sqrt(1-16x^2)-64x^3sqrt(1-16x^2))/1024+C`.

Simplifying this, we get `arcsin(4x)/512+xsqrt(1-16x^2)*(32x^2-1)/256+C`.

Answer 2

`int x^2sqrt(1-16x^2)dx = 1/512(arcsin(4x) - 4x(1-32x^2)sqrt(1-16x^2))+C`

Explanation:

Substitute :

`x=1/4 sint`

`dx = 1/4 cost dt`

as the integrand function is defined only for `x in [-1/4,1/4]`, `t` will vary in the interval `[-pi/2,pi/2]`, so that:

`t=arcsin 4x`

Then:

`int x^2sqrt(1-16x^2)dx = int (1/4sint)^2 sqrt (1-sin^2t) (1/4 cost) dt`

`int x^2sqrt(1-16x^2)dx = 1/64 int sin^2t cost sqrt (1-sin^2t) dt`

Note now that for `t in [-pi/2,pi/2]` we have `cost >0` so that:

`sqrt (1-sin^2t) = cost`

and:

`int x^2sqrt(1-16x^2)dx = 1/64 int sin^2t cos^2t dt`

using now the trigonometric identities:

`2sinalphacosalpha = sin 2alpha`

`sin^2alpha = (1-cos (2 alpha))/2`

we have:

`int x^2sqrt(1-16x^2)dx = 1/256 int sin^2(2t) dt`

and:

`int x^2sqrt(1-16x^2)dx = 1/512 int (1-cos 4t) dt`

`int x^2sqrt(1-16x^2)dx = 1/512(t-sin(4t)/4)+C`

To invert the substitution now expand `sin 4t` using multiple angle formulas:

`sin 4t = 2sin2tcos2t`

`sin 4t = 4 sint cost (cos^2t-sin^2t)`

`sin 4t = 4 sint sqrt(1-sin^2t)(1-2sin^2t)`,

`sin(4t) = 16xsqrt(1-16x^2)(1-32x^2)`

so:

`int x^2sqrt(1-16x^2)dx = 1/512(arcsin(4x) - 4x(1-32x^2)sqrt(1-16x^2))+C`