Hy i have a little problem with solving integral. I have to use new variable as $\displaystyle t=Sin(x)$ $\displaystyle \int \! {Cos(x) \over {Sin(x)^2-3 Sin(x)+2}} \, dx$ But i don't know how to start.
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Originally Posted by PJani Hy i have a little problem with solving integral. I have to use new variable as $\displaystyle t=Sin(x)$ $\displaystyle \int \! {Cos(x) \over {Sin(x)^2-3 Sin(x)+2}} \, dx$ But i don't know how to start. To start set $\displaystyle t= \sin{x}$ . Then: $\displaystyle dt = dx \cos{x}$ and so the integral becomes $\displaystyle \int \frac{dt}{t^2-3t+2}$ To proceed try completing the square in the denominator. This should be enough to get you started. Good luck.
There's no need to complete the square, use that $\displaystyle t^2-3t+2=(t-1)(t-2).$
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