Answer to Question #139778 in Calculus for Moel Tariburu

Question

Evaluate the integral ∫x^3 sin⁡〖x〗dx

Answer 1

$\int x^3sinx\, dx$

$Using\ the\ formula, \int Udv = UV -\int Vdu$

$\begin{aligned} U = x^3 \qquad &; \qquad dv=sinx\,dx \\ \frac{du}{dx}=3x^2\qquad &; \qquad \int dv = \int sinx\, dx\\ du= 3x^2\,dx\qquad &; \qquad V = -cosx \end{aligned}$

$\therefore \int x^3sinx\, dx = (x^3)(-cosx) - \int(-cosx)(3x^2\,dx)$

$\int x^3sinx\, dx = -x^3\,cosx - \int( -3x^2cosx\,dx)$

$\int x^3sinx\, dx = -x^3\,cosx - (-3)\int x^2cosx\,dx$

$\int x^3sinx\, dx = -x^3\,cosx +3 \int x^2cosx\,dx\ --(i)$

$\int x^2cosx\,dx =$

$\begin{aligned} U = x^2\qquad &; \qquad dv = cosx\,dx\\ \frac{du}{dx}= 2x\qquad &; \qquad \int dv = \int cosx\,dx\\ du = 2x\,dx\qquad &; \qquad V = sinx \end{aligned}$

$\therefore \int x^2cosx\, dx = (x^2)(sinx) - \int(sinx)(2x\,dx)$

$\int x^2cosx\, dx = x^2\,sinx - 2\int xsinx\,dx\ ---(ii)$

$\int xsinx\,dx =$

$\begin{aligned} U = x\qquad &; \qquad dv = sinx\,dx\\ \frac{du}{dx}= 1\qquad &; \qquad \int dv = \int sinx\,dx\\ du = dx\qquad &; \qquad V = -cosx \end{aligned}$

$\therefore \int xsinx\, dx = (x)(-cosx) - \int(-cosx)(dx)$

$\int xsinx\, dx = -xcosx + \int cosx\,dx$

$\int xsinx\, dx = -xcosx + sinx$

Substituting the value of $\int xsinx\,dx$ into equation (ii),

$\int x^2cosx\, dx = x^2\,sinx - 2\int xsinx\,dx$

$\int x^2cosx\, dx = x^2\,sinx - 2(-xcosx+sinx)$

$\int x^2cosx\, dx = x^2\,sinx +2xcosx - 2sinx$

Substituting the value of $\int x^2cosx\,dx$ into equation (i), we have

$\int x^3sinx\, dx = -x^3\,cosx +3 \int x^2cosx\,dx$

$\int x^3sinx\, dx = -x^3\,cosx +3 (x^2\,sinx +2xcosx - 2sinx)$

$\int x^3sinx\, dx = -x^3\,cosx +3x^2\,sinx +6xcosx - 6sinx + C$

$\therefore The\ integral\ of\ x^3sinx = -x^3\,cosx +3x^2\,sinx +6xcosx - 6sinx + C$