  # use trig substitution to find integral of 1/[3t^2(6-t^2)^(1/2). that is numerator is 1denominator 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!

check_circleExpert Solution
Step 1

Given:

1/[3t2 (6-t2)1/2]

Step 2

Calculation:

Integrate the given as follows.

Step 3

On further simpli...

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