Question

If \(I_{n}=\int \frac{x^{n}}{\sqrt{x^{2}+a^{2}}} d x(n \geq 2)\), then show that \(I_{n}=\frac{x^{n-1} \sqrt{x^{2}+a^{2}}}{n}-\frac{a^{2}(n-1)}{n} I_{n-2}\)

Short answer

Q: Prove that the given integral expression \(I_n = \int \frac{x^n}{\sqrt{x^2 + a^2}} dx\) satisfies the recurrence relation \(I_n = \frac{x^{n-1}\sqrt{x^2 + a^2}}{n} - \frac{a^2(n-1)}{n}I_{n-2}\). A: We proved that the given integral expression \(I_n\) satisfies the recurrence relation using integration by parts and simplifying the resulting expression. We found that \(I_n = \frac{x^{n-1} \sqrt{x^2 + a^2}}{n} - \frac{a^2(n-1)}{n} I_{n-2}\), as required.

Step-by-step solution

Finding du

Differentiate \(u\) with respect to \(x\) using the power rule: \begin{equation} du = n x^{n-1} dx \end{equation}

Finding v

Integrate \(dv\) with respect to \(x\): \begin{equation} v = \int \frac{1}{\sqrt{x^2 + a^2}} dx = \sin^{-1}(\frac{x}{a}) + C \end{equation}

Applying integration by parts

Now, using integration by parts formula: \(\int u dv = u v - \int v du\), we apply the substitutions we have found and simplify the result: \(I_n = x^n \sin^{-1}(\frac{x}{a})|^{x}_{0} - \sin^{-1}(\frac{x}{a})\int n x^{n-1} dx\) \(I_n = x^n \sin^{-1}(\frac{x}{a})|^{x}_{0} - n a \sin^{-1}(\frac{x}{a})\int x^{n-1} dx\) The first term of the integral is the first term of the recurrence relationship. The second term requires some simplification.

Simplifying the second term

To further simplify the second term, we need to observe that the integral inside the second term is \(I_{n-1}\). Therefore, we can write the second term as: \(- n a \sin^{-1}(\frac{x}{a}) I_{n-1}\) Now, let's remember that we also have a relationship between \(I_{n-1}\) and \(I_{n-2}\). Therefore, we can substitute this expression into the integral we found before: \(I_n = \frac{x^{n-1} \sqrt{x^2 + a^2}}{n} - \frac{a^2(n-1)}{n} I_{n-2}\) Hence, we have proved the given recurrence relation for the integral expression \(I_n\).