Answer to Question #136657 in Calculus for Sagar

Question

Derive the reduction formula integration (x2+a2)n/2 dx = x(x2+a2)/n+1 + na2 /n+1 integration ( x2+a2)n/2-1 dx
Use the formula to integrate [(x2 +a2)n/2-1

Answer 1

$\displaystyle\textsf{Let}\hspace{0.1cm} I_n = \int (x^2 + a^2)^n \hspace{0.1cm} \mathrm{d}x \\ \textsf{Integrating by parts, we have;}\\ I_n = \int (x^2 + a^2)^n \hspace{0.1cm} \mathrm{d}(x) \\ \begin{aligned} I_n &= x(x^2 + a^2)^n - 2n\int x^2(x^2 + a^2)^{n - 1} \hspace{0.1cm} \mathrm{d}x\\ &= x(x^2 + a^2)^n - 2n\int\frac{x^2}{x^2 + a^2}\cdot(x^2 + a^2)^n \mathrm{d}x \\&= x(x^2 + a^2)^n - 2n\int\left(1 - \frac{a^2}{x^2 + a^2}\right)(x^2 + a^2)^n \mathrm{d}x\\&= x(x^2 + a^2)^n - 2n\int (x^2 + a^2)^n - a^2(x^2 + a^2)^{n - 1}\hspace{0.1cm}\mathrm{d}x \end{aligned}\\ I_n = x(x^2 + a^2)^n - 2n(I_n - a^2 I_{n - 1}) \\ I_n (1 + 2n) = x(x^2 + a^2)^n + 2na^2 I_{n - 1}\\ \therefore I_n = \frac{x(x^2 + a^2)^n}{1 + 2n} + \frac{2na^2 I_{n - 1}}{1 + 2n}\\ \textsf{Replacing} \hspace{0.1cm}n \hspace{0.1cm}\textsf{by}\hspace{0.1cm} n/2\\ \therefore I_{n/2} = \frac{x(x^2 + a^2)^n}{1 + n} + \frac{na^2 I_{\frac{n}{2} - 1}}{1 + n}\\ \therefore \int (x^2 + a^2)^{\frac{n}{2}} \hspace{0.1cm} \mathrm{d}x = \frac{x(x^2 + a^2)^n}{1 + n} + \frac{na^2}{1 + n}\int (x^2 + a^2)^{\frac{n}{2} - 1} \hspace{0.1cm} \mathrm{d}x \\ \textsf{Evaluating}\hspace{0.1cm} I_{\frac{n}{2} - 1}, \textsf{We must replace}\hspace{0.1cm} n \hspace{0.1cm}\textsf{with}\hspace{0.1cm} n - 2 \\ \int (x^2 + a^2)^{\frac{n}{2} - 1} \hspace{0.1cm} \mathrm{d}x = \frac{x(x^2 + a^2)^{n - 2}}{n - 1} + \frac{(n - 2)a^2}{n - 1}\int (x^2 + a^2)^{\frac{n}{2} - 2} \hspace{0.1cm} \mathrm{d}x$