Question

If u_n = $\sqrt{(a^{2} - x^{2})}$dx, prove that

u_n = – $\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{(n + 2)}$+ $\frac{(n - 1)}{(n + 2)}$ a² u_{n– 2}

• A. u_n = – $\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{(n + 2)}$ + $\frac{(n - 1)}{(n + 2)}$ a² u_n–2
• B. u_n = $\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{(n + 2)}$ + $\frac{(n - 1)}{(n + 2)}$ a² u_n–2
• C. u_n = $\frac{x^{n - 1}(a^{2} - x^{2})}{(n + 2)}$ + $\frac{(n - 1)}{(n + 2)}$ a² u_n–2
• D. None of these

Answer 1

Answer: (A)

u_n = – $\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{(n + 2)}$ + $\frac{(n - 1)}{(n + 2)}$ a² u_n–2

Answer 2

Q u_n = $\int_{}^{}{x^{n}\sqrt{(a^{2} - x^{2})}}$dx = $\int_{}^{}x^{n - 1}${x $\sqrt{(a^{2} - x^{2})}$} dx

Integrating by parts taking x^n–1 as first function, we have

= x^n–1 $\left\{ \frac{- (a^{2} - x^{2})^{3/2}}{3} \right\}$ +

$\int_{}^{}{(n - 1)}$ $\frac{x^{n - 2}(a^{2} - x^{2})^{3/2}}{3}$dx

Žu_n = –$\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{3}$ + $\frac{(n - 1)}{3}$ $\int_{}^{}x^{n - 2}$ (a² – x²) $\sqrt{(a^{2} - x^{2})}$dx

Žu_n = –$\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{3}$ + $\frac{(n - 1)a^{2}}{3}$

u_{n – 2} – $\frac{(n - 1)}{3}$u_n

Ž $\frac{(n + 2)}{3}$u_n = –$\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{3}$ + $\frac{(n - 1)a^{2}u_{n - 2}}{3}$

Ž u_n = – $\frac{x^{n - 1}(a^{2} - x^{2})^{3/2}}{(n + 2)}$ + $\frac{(n - 1)}{(n + 2)}$ a² u_{n– 2}