- #1

- 13

- 1

Use the substitution x = 3sint to show that

3

[inte]x^2[squ](9-x^2) dx = (81/16)pi

0

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- Thread starter Einstein
- Start date

- #1

- 13

- 1

Use the substitution x = 3sint to show that

3

[inte]x^2[squ](9-x^2) dx = (81/16)pi

0

- #2

- 54

- 1

- #3

- 44

- 0

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 ya without actually performing the written instructions. Hope this helps in future endeavors as well as the current problem.

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