- #1
- 13
- 1
How do you do this question, I've spent hours figuring it out:
Use the substitution x = 3sint to show that
3
[inte]x^2[squ](9-x^2) dx = (81/16)pi
0
Use the substitution x = 3sint to show that
3
[inte]x^2[squ](9-x^2) dx = (81/16)pi
0
Integration by substitution question
- Thread starter Einstein
- Start date
- #2
- 54
- 1
Have you got it anywhere near [inte] 81 (sint)^2 * (cost)^2 dt from 0 to [pi]/2 yet? You can make a substitution from there. If you haven't, have you forgot to substitute dx for 3cost dt, and changed the limits?
- #3
- 44
- 0
This is just a simple Trig sub problem. x=3sin(t), therefore x^2=9sin^(t), and 9-x^2=9-9sin^2(t) or 9-x^2=9cos^2(t), and dx=3cos(t)so the problem becomes:
3
[inte] 9sin^2(t)*3cos(t)*3cos(t)dt
0
the factor out the constants and then sub sin^2(t) as (1-cos^2(t)), then distribute the other cos^2(t), and bust out an integration table for cos^2(t) and cos^4(t).. that's about all I can tell you without actually performing the written instructions. Hope this helps in future endeavors as well as the current problem.
3
[inte] 9sin^2(t)*3cos(t)*3cos(t)dt
0
the factor out the constants and then sub sin^2(t) as (1-cos^2(t)), then distribute the other cos^2(t), and bust out an integration table for cos^2(t) and cos^4(t).. that's about all I can tell you without actually performing the written instructions. Hope this helps in future endeavors as well as the current problem.