Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 24

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Evaluate $\displaystyle \int^\pi_0 x^2 \cos x dx$ by using Integration by parts.

If we let $u = x^3$ and $dv = \cos x dx$, then

$du = 3x^2 dx$ and $\displaystyle v = \int \cos x dx = \sin x$

So,

$\displaystyle \int^\pi_0 x^3 \cos x dx = uv - \int v du = x^3 \sin x - \int 3x^2 \sin x dx$

To evaluate $\displaystyle \int 3x^2 \sin x dx$, we must also use integration by parts so

if we let $u_1 = 3x^2$ and $dv_1 = \sin x dx$, then

$du_1 = 6x dx$ and $\displaystyle v_1 = \int \sin x dx = - \cos x$

So,

$ \begin{equation} \begin{aligned} \int 3x^2 \sin x dx = u_1 v_1 - \int v_1 du_1 &= -3x^2 \cos x - \int - 6x \cos x dx\\ \\ &= -3x^2 \cos x + 6 \int x \cos x dx \end{aligned} \end{equation} $

To evaluate $\displaystyle \int x \cos dx$, we let

$ \begin{equation} \begin{aligned} u_2 &= x \text{ and } dv_2 = \cos x dx, \text{ then}\\ \\ du_2 &= dx \text{ and } v_2 = \sin x \end{aligned} \end{equation} $

$ \begin{equation} \begin{aligned} \text{so } \int x \cos x dx = u_2 v_2 - \int v_2 du_2 &= x \sin x - \int \sin x dx\\ \\ &= x \sin x - (-\cos x)\\ \\ &= x \sin x + \cos x \end{aligned} \end{equation} $

Going back to the first equation,

$ \begin{equation} \begin{aligned} \int^\pi_0 x^3 \cos x dx &= x^3 \sin x - \int 3x^2 \sin x dx\\ \\ &= x^3 \sin x - \left[ -3x^2 \cos x + 6 \left( x \sin x + \cos x \right)\right]\\ \\ &= x^3 \sin x + 3x^2 \cos x - 6 x \sin x - 6 \cos x \end{aligned} \end{equation} $

Evaluating from 0 to $\pi$,

$ = - 3 \pi^2 + 12$