Question

(a) Let In = \int _0^1x^n\sqrt{1-x^2\:dx}, n ∈ N. Show that (n + 2)In = (n − 1)In−2, n ≥ 2.

(b) If g = sin(sin x), prove that \frac{d^2y}{dx^2}+tan\:x\frac{dy}{dx}+ycosx^2= 0.

Answer 1

(a)

$$I_n=\displaystyle\int_{0}^{1}x^n\sqrt{1-x^2}dx$$

$$\int x^n\sqrt{1-x^2}dx$$

$$\int udv=uv-\int vdu$$

$$u=x^{n-1}, du=(n-1)x^{n-2}dx$$

$$dv=x\sqrt{1-x^2}dx, v=\int x\sqrt{1-x^2}dx$$

$$=-\dfrac{1}{3}(1-x^2)^{{3 \over 2}}$$

$$\int x^n\sqrt{1-x^2}dx=x^{n-1}\big(-\dfrac{1}{3}(1-x^2)^{{3 \over 2}}\big)$$

$$+\dfrac{n-1}{3}\int x^{n-2}(1-x^2)\sqrt{1-x^2}dx$$

$$=x^{n-1}\big(-\dfrac{1}{3}(1-x^2)^{{3 \over 2}}\big)+\dfrac{n-1}{3}\int x^{n-2}\sqrt{1-x^2}dx$$

$$-\dfrac{n-1}{3}\int x^{n}\sqrt{1-x^2}dx$$

$$(n+2)\int x^n\sqrt{1-x^2}dx=x^{n-1}(1-x^2)^{{3 \over 2}}$$

$$+(n-1)\int x^{n-2}\sqrt{1-x^2}dx$$

$$(n+2)\displaystyle\int_{0}^{1}x^n\sqrt{1-x^2}dx$$

$$=\big[x^{n-1}(1-x^2)^{{3 \over 2}}\big]\begin{matrix} 1 \\ 0 \end{matrix}+(n-1)\displaystyle\int_{0}^{1}x^{n-2}\sqrt{1-x^2}dx$$

$$(n+2)\displaystyle\int_{0}^{1}x^n\sqrt{1-x^2}dx$$

$$=(0-0)+(n-1)\displaystyle\int_{0}^{1}x^{n-2}\sqrt{1-x^2}dx$$

Hence

$$(n+2)I_n=(n-1)I_{n-2}, \ \ n\geq2$$

(b)

$$g=\sin(\sin x)$$

$$\dfrac{dg}{dx}=\cos(\sin x)\cdot\cos x$$

$$\dfrac{d^2 g}{dx^2 }=-\sin(\sin x)\cdot\cos^2 x-\cos(\sin x)\cdot\sin x$$

$$\dfrac{d^2 g}{dx^2 }+\tan x\cdot\dfrac{d g}{dx }+g\cdot \cos^2x$$

$$=-\sin(\sin x)\cdot\cos^2 x-\cos(\sin x)\cdot\sin x$$

$$+\tan x\cdot\cos(\sin x)\cdot\cos x$$

$$+\sin(\sin x)\cdot\cos^2x$$

$$=0$$