# use trig substitution to find integral of 1/[3t^2(6-t^2)^(1/2).   that is numerator is 1 denominator is 3 t-squared times the square root of 6-t-squared.   I tried letting t=square root of six times cosx. I worked it to the point where I have 1 over 18 times square root 6 times cos^2(x) sin(x).  then i brought those to numerator in the form of sec^2(x)csc(x) with 1 over 18 square root 6 going outside of the integral.  If I did this correct up to this point, how can I solve if it is in terms of sec and csc?  Thanks!

Question

use trig substitution to find integral of 1/[3t^2(6-t^2)^(1/2).

that is numerator is 1

denominator is 3 t-squared times the square root of 6-t-squared.

I tried letting t=square root of six times cosx.

I worked it to the point where I have 1 over 18 times square root 6 times cos^2(x) sin(x).  then i brought those to numerator in the form of sec^2(x)csc(x) with 1 over 18 square root 6 going outside of the integral.  If I did this correct up to this point, how can I solve if it is in terms of sec and csc?  Thanks!

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MathCalculus