# Thread: how would I Evaluate this??

1. ## how would I Evaluate this??

∫[(sinxdx)/(cos^2x + cos x -2)]

what i did was integration by parts but it is wrong.....and i couldn't even finish it

=∫sinxdx/(cosx-1)(cosx+2)

sinx/cos^2+cosx-2 = A/cosx-1 + B/cosx +2
sinx = Acosx + 2A + Bcosx -B
sinx = cosx(A+B) + (2A-B)

2A-B=0

2. Originally Posted by Yello
∫[(sinxdx)/(cos^2x + cos x -2)]

what i did was integration by parts but it is wrong.....and i couldn't even finish it

=∫sinxdx/(cosx-1)(cosx+2)

sinx/cos^2+cosx-2 = A/cosx-1 + B/cosx +2
sinx = Acosx + 2A + Bcosx -B
sinx = cosx(A+B) + (2A-B)

2A-B=0
$\int{\frac{\sin{x}}{\cos^2{x} + \cos{x} - 2}\,dx}$.

You need to use a $u$ substitution:

$u = \cos{x}$ so that $\frac{du}{dx} = -\sin{x}$.

$\int{\frac{\sin{x}}{\cos^2{x} + \cos{x} - 2}\,dx} = -\int{\frac{1}{\cos^2{x} + \cos{x} - 2}(-\sin{x})\,dx}$

$= -\int{\left(\frac{1}{u^2 + u - 2}\right)\frac{du}{dx}\,dx}$

$= -\int{\frac{1}{(u + 2)(u - 1)}\,du}$.

Now use a partial fractions decomposition.