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
Asked Mar 18, 2019

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.

fullscreen
Step 3

On further simpli...

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour*

See Solution
*Response times may vary by subject and question
Tagged in

Math

Calculus

Integration