Question

Verify the value of the definite integral, where \(n\) is a positive integer. $$ \int_{-\pi}^{\pi} x^{2} \cos n x d x=\frac{(-1)^{n} 4 \pi}{n^{2}} $$

Short answer

The definite integral value \(\int_{-\pi}^{\pi} x^{2} \cos n x dx\) equals \(\frac{(-1)^{n} 4 \pi}{n^{2}}\), thus verifying the original equation.

Step-by-step solution

Understand the integration by parts formula

The integration by parts formula is given by \(\int u v dx = u \int v dx - \int u' (\int v dx) dx\). We can choose \(u=x^2\) and \(v=\cos nx\) to proceed in order to calculate the integral.

Differentiating and integrating \(u\) and \(v\)

Differentiate \(u\) and integrate \(v\) respectively to obtain the values of \(du\) and \(v\) which will be substituted in the above equation, which gives \(du = 2xdx\) and \(\int v dx = \int \cos nx dx = \frac{1}{n}\sin nx\).

Substitute \(u\), \(v\), \(du\) and \(\int v dx\) into the formula and find the integral terms

The integration by parts formula \(u \int v dx - \int u' (\int v dx) dx\) becomes \(x^2 \frac{1}{n}\sin nx - \int 2x (\frac{1}{n}\sin nx) dx\). Evaluate the integral from \(-\pi\) to \(\pi\) and observe the property of the sine function over the period \(-\pi\) to \(\pi\).

Evaluate the first term

The first term becomes zero because the integration from \(-\pi\) to \(\pi\) of \(x^2 \sin nx\), is zero as \(\sin nx\) is an odd function and the product of even function \(x^2\) and odd function \(\sin nx\) results in an odd function and integral of an odd function over a symmetric interval is 0.

Evaluate the second term

The second term can be calculated using integration by parts twice and observing that \(\int_{-\pi}^{\pi}x \sin nx dx = 0\) (due to the sine function being odd and \(x\) also being odd, the product becomes even but the integral over a symmetric interval of such a function equals 0) and \(\int_{-\pi}^{\pi} \sin nx dx = 0\) (the integral of the sine function from \(-\pi\) to \(\pi\) equals 0).

Compute the integral

Computing the integral gives the result \(\int_{-\pi}^{\pi} x^{2} \cos n x dx=\frac{(-1)^{n} 4 \pi}{n^{2}}\) as expected.