Integral from pi/2 to -pi/2 (x^2 sinx) / (1 + x^6) dx, and
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Re: Integral help.
okay, so for the first one if i let u = a^2-x^2, than du would be 2a-2x dx ......
which i don't understand how that would help because than i would need to put 2a-2x in there in order to change from dx to du?? sorry i'm confused..
the 2nd one, i understand if it's an odd function than f(x)dx = 0 right? but i don't know where to go from there. would i change the equation to be x^2sinx = 1+x^6 and integrate each side on their own? .. can you tell my prof doesn't ever explain things? lol thanks a lot for your time
okay, so for the first one if i let u = a^2-x^2, than du would be 2a-2x dx ......
which i don't understand how that would help because than i would need to put 2a-2x in there in order to change from dx to du?? sorry i'm confused..
the 2nd one, i understand if it's an odd function than f(x)dx = 0 right? but i don't know where to go from there. would i change the equation to be x^2sinx = 1+x^6 and integrate each side on their own? .. can you tell my prof doesn't ever explain things? lol thanks a lot for your time
Re: Integral help.
The first one is a booger to integrate. Actually, it's not easily done by elementary means. I think it's a matter of observation. You are expected to notice it's an odd function and that it is 0. You are not expected to actually integrate it by brute force.
For the second one, \(\displaystyle \int_{a}^{0}x\sqrt{a^{2}-x^{2}}dx\)
you could also let \(\displaystyle x=asin({\theta}), \;\ dx=acos({\theta})d{\theta}\)
The first one is a booger to integrate. Actually, it's not easily done by elementary means. I think it's a matter of observation. You are expected to notice it's an odd function and that it is 0. You are not expected to actually integrate it by brute force.
For the second one, \(\displaystyle \int_{a}^{0}x\sqrt{a^{2}-x^{2}}dx\)
you could also let \(\displaystyle x=asin({\theta}), \;\ dx=acos({\theta})d{\theta}\)
skeeter
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\(\displaystyle \int_a^0 x\sqrt{a^2 - x^2} dx\)
\(\displaystyle u = a^2 - x^2\)
a is a constant! ...
\(\displaystyle du = -2x \, dx\)
\(\displaystyle -\frac{1}{2} \int_a^0 -2x\sqrt{a^2 - x^2} dx\)
\(\displaystyle -\frac{1}{2} \int_0^{a^2} \sqrt{u} du\)
finish up.
\(\displaystyle u = a^2 - x^2\)
a is a constant! ...
\(\displaystyle du = -2x \, dx\)
\(\displaystyle -\frac{1}{2} \int_a^0 -2x\sqrt{a^2 - x^2} dx\)
\(\displaystyle -\frac{1}{2} \int_0^{a^2} \sqrt{u} du\)
finish up.