If I_m = int x^m cos n x , dx = g(x) - frac{m | vedclass

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(D) We are given $I_m = \int x^m \cos n x \, dx$. Using integration by parts, let $u = x^m$ and $dv = \cos n x \, dx$. Then $du = m x^{m-1} \, dx$ and

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$\frac{x^m \sin n x}{n} + \frac{m}{n^2} x^{m-1} \cos n x$

Vedclass · Answer

(D) We are given $I_m = \int x^m \cos n x \, dx$. Using integration by parts, let $u = x^m$ and $dv = \cos n x \, dx$. Then $du = m x^{m-1} \, dx$ and $v = \frac{\sin n x}{n}$.$I_m = \frac{x^m \sin n x}{n} - \int \frac{m x^{m-1} \sin n x}{n} \, dx = \frac{x^m \sin n x}{n} - \frac{m}{n} \int x^{m-1} \sin n x \, dx$.Now, integrate $\int x^{m-1} \sin n x \, dx$ by parts again. Let $u = x^{m-1}$ and $dv = \sin n x \, dx$. Then $du = (m-1) x^{m-2} \, dx$ and $v = -\frac{\cos n x}{n}$.$\int x^{m-1} \sin n x \, dx = -\frac{x^{m-1} \cos n x}{n} + \int \frac{(m-1) x^{m-2} \cos n x}{n} \, dx$.Substituting this back into the expression for $I_m$:$I_m = \frac{x^m \sin n x}{n} - \frac{m}{n} \left( -\frac{x^{m-1} \cos n x}{n} + \frac{m-1}{n} \int x^{m-2} \cos n x \, dx \right)$.$I_m = \frac{x^m \sin n x}{n} + \frac{m x^{m-1} \cos n x}{n^2} - \frac{m(m-1)}{n^2} I_{m-2}$.Comparing this with the given equation $I_m = g(x) - \frac{m(m-1)}{n^2} I_{m-2}$, we get $g(x) = \frac{x^m \sin n x}{n} + \frac{m x^{m-1} \cos n x}{n^2}$.

If $I_m = \int x^m \cos n x \, dx = g(x) - \frac{m(m-1)}{n^2} I_{m-2}$, then $g(x) = $

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