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I rewrite to cos^6x/sin^3x dx and let u = sinx but when I'm trying to rewrite integral, what should I do with the ^6?

Thanks!

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- Thread starter anderma8
- Start date

- #1

- 35

- 0

I rewrite to cos^6x/sin^3x dx and let u = sinx but when I'm trying to rewrite integral, what should I do with the ^6?

Thanks!

- #2

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- #3

HallsofIvy

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[tex]\int \frac{cos^6 t}{sin^3 t}dt= \frac{cos ^6 t sin t}{sin^4 t}dt[/tex]

[tex]= \int \frac{cos^6 t}{(1- cos^2 t)^2}(sin t dt)[/tex]

so that the substitution x= cos t gives

[tex]-\int\frac{x^6}{(1-x^2)^2}dx[/tex]

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