# Integrate w/negative exponent

1. Apr 15, 2007

### anderma8

I need a little 'suggestion' as to how to integrate cos^6x sin^-3x dx.

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. Apr 15, 2007

### anderma8

oh wait, re-reading the problem and I think what I'm interpreting as a negative is the dot in the i in sin.... I'm going to work the problem as cos^6x sin^3x dx and I know how to work this... sorry about that!

3. Apr 16, 2007

### HallsofIvy

Staff Emeritus
For future reference, if you did have an odd power of sine in the denominator, multiply both numerator and denominator by sine:
$$\int \frac{cos^6 t}{sin^3 t}dt= \frac{cos ^6 t sin t}{sin^4 t}dt$$
$$= \int \frac{cos^6 t}{(1- cos^2 t)^2}(sin t dt)$$
so that the substitution x= cos t gives
$$-\int\frac{x^6}{(1-x^2)^2}dx$$