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OpenStudy (anonymous):
Find the integral of arcsec(x)/(x^6). I arrived at the integral 1/(sec^5) with integration by parts and trig substitution but now I don't know what to do.
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OpenStudy (anonymous):
That's rough, I really don't how to to figure that one out. Sorry.
OpenStudy (anonymous):
now, 1/(secx^5)=(1/secx)^5=cosx^5 so the integral of 1/(sec^5) is the same as the integral of cos^5 =int (cosx^4*cosx) dx =int [[(1-sinx^2)^2]*cosx] dx use substitution again, let u=sinx => du=cosdx then we have =int [(1-u^2)^2)du=int(1-2u^2+u^4)du =u-2u^3/3+u^5/5 =sinx-2sinx^3/3+sinx^5/5
OpenStudy (anonymous):
thank you very much!
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