Question

Evaluate question 3 only ​

Answer 1

Step-by-step explanation:

(4²-x²)³/²

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

Answer 2

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 $\sqrt{x^2} = |x|$ for all real $x$)

So, the integral we want transforms to

$\displaystyle \int (16 - x^2)^{3/2} \, dx = 64 \int \cos^3(y) \times 4\cos(y) \, dy = 256 \int \cos^4(y) \, dy$

Expand the integrand using the identity

$\cos^2(x) = \dfrac{1+\cos(2x)}2$

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,

$\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \dfrac{x^2}{16}} = \dfrac{\sqrt{16-x^2}}4$

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

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