- #1

CellCoree

- 42

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what would be the appropriate trigonometic substition for the following:

1.) [tex]\int(7x^2-7)^{3/2}) dx [/tex]

2.) [tex] \int \frac{x^2}{\sqrt{(7x^2 + 8)}} [/tex]

ok i have a book with a list of trigonometric functions in front of me, but i don't get it. i don't see anything that i can sub in for.

ok i kinda get the first one...

[tex]7(x^2-1)^{3/2}[/tex]

[tex]sin^2(\theta) + cos^2(\theta) = 1/[/tex] so...

divided by [tex]cos^2(\theta)[/tex] and get [tex]tan^2(\theta)[/tex]

[tex]7(x^2 - tan^2(\theta))^{3/2}[/tex]

[tex]7(x^3 - tan^3(\theta))[/tex]

[tex]7x^3 - 7tan^3(\theta)[/tex]

[tex]7x^3=7tan^3(\theta)[/tex]

divide by 7... (lol, don't think i can do this)

[tex]x^3 = tan^3(\theta)[/tex]

cube root...

[tex]x= tan(\theta)[/tex] is that right?

1.) [tex]\int(7x^2-7)^{3/2}) dx [/tex]

2.) [tex] \int \frac{x^2}{\sqrt{(7x^2 + 8)}} [/tex]

ok i have a book with a list of trigonometric functions in front of me, but i don't get it. i don't see anything that i can sub in for.

ok i kinda get the first one...

[tex]7(x^2-1)^{3/2}[/tex]

[tex]sin^2(\theta) + cos^2(\theta) = 1/[/tex] so...

divided by [tex]cos^2(\theta)[/tex] and get [tex]tan^2(\theta)[/tex]

[tex]7(x^2 - tan^2(\theta))^{3/2}[/tex]

[tex]7(x^3 - tan^3(\theta))[/tex]

[tex]7x^3 - 7tan^3(\theta)[/tex]

[tex]7x^3=7tan^3(\theta)[/tex]

divide by 7... (lol, don't think i can do this)

[tex]x^3 = tan^3(\theta)[/tex]

cube root...

[tex]x= tan(\theta)[/tex] is that right?

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