The integral of xarccos(x)dx
It's supposed to be solved using a table of integrals...however, my prof. wants us to solve all our HW probs NOT using a table of integrals.
Any help is appreciated, even any hints about how to go about solving w/o integral table would be great! Thanks
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OpenStudy (anonymous):
by parts
OpenStudy (anonymous):
u = cos^-1(x)
dv = x dx
OpenStudy (anonymous):
du = -1/ sqrt(1-x^2) dx
v = (1/2)x^2
OpenStudy (anonymous):
Thanks! I was leaning towards parts, but was second guessing.
OpenStudy (anonymous):
it should be easy
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OpenStudy (anonymous):
v du is just an algebraic function
OpenStudy (anonymous):
easy to integrate
OpenStudy (anonymous):
wait
OpenStudy (anonymous):
Cool. I'll give it a shot
OpenStudy (anonymous):
u need trig sub
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OpenStudy (anonymous):
instead of parts altogether? Or in addition to?
OpenStudy (anonymous):
no u need parts
OpenStudy (anonymous):
then integration by parts says its
uv - integral v du
OpenStudy (anonymous):
v du is something like x^2 / sqrt(1-x^2) with some constant floating around
OpenStudy (anonymous):
Oh i see where it's going to need trig substitution
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