integral of (x^2*sin x)/(1+x^6) on the interval [-pi/2, pi/2]
Supposed to use substitution somehow, but can't figure out how. Any help would be appreciated.
2006-09-23 00:12:39 · 4 answers · asked by Voodoo6969_98 2 in Science & Mathematics ➔ Mathematics
It is a "trick" question to see if you understand properties of integrals.
f(x)=(x^2*sin(x))/(1+x^6) is an odd function so the integral of f(x) from -pi/2 to pi/2 is 0.
(In terms of area, the area above the graph and below the x-axs from -pi/2 to 0 is the same as the area below the graph and above the x-axis from 0 to pi/2.)
PS f(x) is odd because x^2/(1+x^6) is an even function and sin(x) is an odd function; the product of an even and an odd function is odd.
2006-09-23 00:26:48 · answer #1 · answered by Anonymous · 3⤊ 0⤋
To use substitution practice is required ;substitution vary from question to question and by a particular substitution question becomes easy . try x square = p-1
2006-09-23 07:27:13 · answer #2 · answered by deepak57 7 · 0⤊ 1⤋
check whether it is a odd function or neven one...then use the approriate formula
2006-09-23 07:22:08 · answer #3 · answered by Mol 2 · 0⤊ 1⤋
use the conjuctive method
2006-09-23 07:15:39 · answer #4 · answered by Techno 2 · 0⤊ 2⤋