Let $I_n = \int_0^1 x^n \sqrt{1 - x^2} \, dx$, for $n = 0, 1, 2, \ldots$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2017 - Paper 1

We shall use integration by parts for:

$$I_n = \int_0^1 x^n \sqrt{1 - x^2} , dx.$$$$ Let ( u = \sqrt{1 - x^2} ) and ( dv = x^n , dx ), then:

$du = -\frac{x}{\sqrt{1 - x^2}} \, dx, \, v = \frac{x^{n+1}}{n + 1}.$

Then,

$I_n = \left[ u v \right]_0^1 - \int_0^1 v \, du$ $= \left[ \frac{x^{n+1} \sqrt{1 - x^2}}{n + 1} \right]_0^1 - \int_0^1 \frac{x^{n + 1}}{n + 1} \left(-\frac{x}{\sqrt{1 - x^2}} \right) dx$ $= 0 + \frac{1}{2(n + 1)} \int_0^1 x^{n + 2} (1 - x^2)^{-1/2} dx$ Finally, simplifying gives:

$I_n = \frac{(n - 1)}{(n + 2)} I_{n - 2}.$

Using the recursive formula:

$I_5 = \frac{(5 - 1)}{(5 + 2)} I_{3} = \frac{4}{7} I_{3}.$

We can repeat this process: $I_3 = \frac{(3 - 1)}{(3 + 2)} I_{1} = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}.$

So, $I_5 = \frac{4}{7} \cdot \frac{2}{15} = \frac{8}{105}.$

To find the equation of the tangent, we differentiate $\sqrt{x} + \sqrt{y} = \sqrt{a}$ implicitly:

$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0,$

which gives: $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}.$

At point P(c, d): $\frac{dy}{dx} = -\frac{\sqrt{d}}{\sqrt{c}}.$

Using point-slope form:

$y - d = -\frac{\sqrt{d}}{\sqrt{c}}(x - c).$ Rearranging leads us to: $\sqrt{c} + x \sqrt{d} = d + c \sqrt{a}.$

Let A be where the tangent meets the x-axis and B where it meets the y-axis. The intercept on the x-axis can be found by setting $y=0$ in the tangent equation, and similarly for the y-intercept.

From geometry, the lengths OA and OB can be calculated and adding them gives:

$OA + OB = c + d = a,$

showing the required result.