- #1

- 223

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By what rules is the integral of

[tex]\frac{cos(t)}{1+9sin^2(t)}=\frac{1}{3}tan^{-1}(sin(3t))[/tex]

I know this is right, but I have no idea why and can't find any trig identities to help. Thanks.

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- Thread starter americanforest
- Start date

- #1

- 223

- 0

By what rules is the integral of

[tex]\frac{cos(t)}{1+9sin^2(t)}=\frac{1}{3}tan^{-1}(sin(3t))[/tex]

I know this is right, but I have no idea why and can't find any trig identities to help. Thanks.

- #2

- 1,235

- 1

Let [tex] u = 3\sin t [/tex]

and use the fact that [tex]\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C [/tex]

and use the fact that [tex]\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C [/tex]

Last edited:

- #3

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gotcha, thanks

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