Integration — practice questions

(a) Deriving the reduction formula: We use integration by parts on $I_n = \int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \sin^{n-1} x \sin x \, dx$. Let $u = \sin^{n-1} x$ and $\frac{dv}{dx} = \sin x$. Then $\frac{du}{dx} = (n-1)\sin^{n-2} x \cos x$ and $v = -\cos x$.

Applying the integration by parts formula: $I_n = [uv]_0^{\pi/2} - \int_0^{\pi/2} v \frac{du}{dx} \, dx$ $I_n = [-\cos x \sin^{n-1} x]_0^{\pi/2} - \int_0^{\pi/2} (-\cos x)(n-1)\sin^{n-2} x \cos x \, dx$

The first term is $[(-\cos(\pi/2) \sin^{n-1}(\pi/2)) - (-\cos(0) \sin^{n-1}(0))] = [0 - 0] = 0$ for $n \ge 2$.

$I_n = (n-1) \int_0^{\pi/2} \cos^2 x \sin^{n-2} x \, dx$ Using the identity $\cos^2 x = 1 - \sin^2 x$: $I_n = (n-1) \int_0^{\pi/2} (1 - \sin^2 x) \sin^{n-2} x \, dx$ $I_n = (n-1) \left( \int_0^{\pi/2} \sin^{n-2} x \, dx - \int_0^{\pi/2} \sin^n x \, dx \right)$ $I_n = (n-1)(I_{n-2} - I_n)$ $I_n = (n-1)I_{n-2} - (n-1)I_n$ $I_n + (n-1)I_n = (n-1)I_{n-2}$ $nI_n = (n-1)I_{n-2}$, as required.

(b) Calculating $I_5$: Using the formula $I_n = \frac{n-1}{n}I_{n-2}$: $I_5 = \frac{4}{5} I_3$ $I_3 = \frac{2}{3} I_1$

We need to calculate the base case $I_1$: $I_1 = \int_0^{\pi/2} \sin x \, dx = [-\cos x]_0^{\pi/2} = (-\cos(\pi/2)) - (-\cos(0)) = 0 - (-1) = 1$.

Now substitute back: $I_3 = \frac{2}{3} \times 1 = \frac{2}{3}$ $I_5 = \frac{4}{5} \times I_3 = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15}$.

First, we need to find the derivatives with respect to $t$: $\frac{dx}{dt} = 3t^2 - 3$ $\frac{dy}{dt} = 6t$

Next, we set up the expression for the integrand in the arc length formula, $s = \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$.

$(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = (3t^2 - 3)^2 + (6t)^2$ $= (9t^4 - 18t^2 + 9) + 36t^2$ $= 9t^4 + 18t^2 + 9$

This expression can be factorised. Notice it's a perfect square: $= 9(t^4 + 2t^2 + 1)$ $= 9(t^2 + 1)^2$

Now, take the square root: $\sqrt{9(t^2 + 1)^2} = 3(t^2 + 1)$ (since $t^2+1$ is always positive).

Now we can set up and evaluate the integral for the arc length from $t=0$ to $t=2$: $s = \int_0^2 3(t^2 + 1) \, dt$ $s = 3 \int_0^2 (t^2 + 1) \, dt$ $s = 3 \left[ \frac{t^3}{3} + t \right]_0^2$ $s = 3 \left( (\frac{2^3}{3} + 2) - (\frac{0^3}{3} + 0) \right)$ $s = 3 \left( \frac{8}{3} + 2 \right)$ $s = 3 \left( \frac{8}{3} + \frac{6}{3} \right) = 3 \left( \frac{14}{3} \right) = 14$.

The exact length of the curve is 14 units.