Question

Evaluate question 3 only ​

Answer 1

Substitute
`x = 4 \sin(y)`, so that
`dx = 4\cos(y)\,dy`. Part of the integrand reduces to

`16 - x^2 = 16 - (4\sin(y))^2 = 16 - 16 \sin^2(y) = 16 (1 - \sin^2(y)) = 16 \cos^2(y)`

Note that we want this substitution to be reversible, so we tacitly assume
`-\frac\pi2\le y\le \frac\pi2`. Then
`\cos(y)\ge0`, and

`(16-x^2)^(3/2) = 16^(3/2) \left(\cos^2(y)\right)^(3/2) = 64 |\cos(y)|^3 = 64 \cos^3(y)`

(since
`√(x^2) = |x|` for all real
`x`)

So, the integral we want transforms to

`\displaystyle \int (16 - x^2)^(3/2) \, dx = 64 \int \cos^3(y) * 4\cos(y) \, dy = 256 \int \cos^4(y) \, dy`

Expand the integrand using the identity

`\cos^2(x) = \frac{1+\cos(2x)}2`

to write

`\displaystyle \int (16 - x^2)^(3/2) \, dx = 256 \int \left(\frac{1 + \cos(2y)}2\right)^2 \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \cos^2(2y)) \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \frac{1 + \cos(4y)}2\right) \, dy \\\\ = 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy`

Now integrate to get

`\displaystyle 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy = 32 \left(3y + 2 \sin(2y) + \frac14 \sin(4y)\right) + C \\\\ = 96 y + 64 \sin(2y) + 8 \sin(4y) + C`

Recall the double angle identity,

`\sin(2y) = 2 \sin(y) \cos(y)`

`\implies \sin(4y) = 2 \sin(2y) \cos(2y) = 4 \sin(y) \cos(y) (\cos^2(y) - \sin^2(y))`

By the Pythagorean identity,

`\cos(y) = √(1 - \sin^2(y)) = \sqrt{1 - (x^2)/(16)} = \frac{√(16-x^2)}4`

Finally, put the result back in terms of
`x`.

`\displaystyle \int (16 - x^2)^(3/2) \, dx \\\\ = 96 \sin^(-1)\left(\frac x4\right) + 128 \frac x4 \frac{√(16-x^2)}4 + 32 \frac x4 \frac{√(16-x^2)}4 \left((16-x^2)/(16) - (x^2)/(16)\right) + C \\\\ = 96 \sin^(-1)\left(\frac x4\right) + 8 x √(16 - x^2) + \frac14 x √(16 - x^2) (8 - x^2) + C \\\\ = \boxed{96 \sin^(-1)\left(\frac x4\right) + \frac14 x √(16 - x^2) \left(40 - x^2\right) + C}`

Answer 2

Explanation:

(4²-x²)³/²

or,(4+X) (4-X)