Integration By Parts - Another Problem

[RedBarchetta]

Homework Statement

$$\int {\frac{{\cos ydy}}<br /> {{\sin ^2 y + \sin y - 6}} }$$

The Attempt at a Solution

$$\int {\frac{{\cos ydy}}<br /> {{\sin ^2 y + \sin y - 6}} = } \int {\frac{{\cos ydy}}<br /> {{(\sin y - 2)(\sin y + 3)}}}$$

Now I attempt to split this into partial fractions:

$$\begin{gathered}<br /> \frac{{\cos y}}<br /> {{(\sin y - 2)(\sin y + 3)}} = \frac{A}<br /> {{\sin y - 2}} + \frac{B}<br /> {{\sin y + 3}} \hfill \\<br /> \cos y = A(\sin y + 3) + B(\sin y - 2) \hfill \\<br /> \cos y = A\sin y + 3A + B\sin y - 2B \hfill \\<br /> \cos y = (A + B)\sin y + 3A - 2B \hfill \\ <br /> \end{gathered}$$

...from here I'm not sure what to do to solve for the coefficients. I can't see any trig identities that would help either. How would you solve for these?

Thank you.

 

[tiny-tim]

$$\int {\frac{{\cos ydy}}<br /> {{\sin ^2 y + \sin y - 6}} }$$

Oh, RedBarchetta!

This is screaming out for a substitution!

 

[RedBarchetta]

Oh, RedBarchetta!

This is screaming out for a substitution!

You're right. It just took me a while to notice that. It definitely decreases the difficulty level.

Thanks.