Question

Integrate the following by expanding the partial fractions:
∫[cos(x)]/[sin^2(x)-sin(x)-2] dx

Answer 1

\[\frac{\cos (x)}{(\sin (x)-2) (\sin (x)+1)}\]

Answer 2

cos(x)/[(sin(x)-2)(sin(x)+1)]
A B
---- + ----
sin(x)-2 sin(x)+1

A=(1/3)cos(x)
B=-(1/3)cos(x)

Answer 3

Let u = sinx
therefore du = cosx dx
Substituting these two into your integral makes it easier to solve so now you have:
\[\int\limits_{}^{} du / (u + 1)(u - 2)du\]

Answer 4

so u need to integrate:
\[(1/3)\int\limits_{}^{}\cos(x)/(\sin(x)-2)-\cos(x)/(\sin(x)+1)\]

Answer 5

1 / (u +1)(u - 2) = A / (u + 1) + B / (u - 2)

Answer 6

Multiply both sides by the lowest common denominator which is (u+ 1)(u - 2)

Answer 7

Therefore: 1 = A(u - 2) + B(u + 1)
If u = 2 ; 1 = 3B, B = 1/3
if u = -1; 1 = -3A, A = -1/3

Answer 8

So now you have\[1/3\int\limits_{}^{} 1 / [ (u - 2) - 1 / (u + 1) ] du\]

Answer 9

1/3 [ ln|u - 2| - ln | u + 2| ] + C

Answer 10

Sorry previous post should be u + 1
1/3 ln|(u - 2)/(u + 1)|

Answer 11

NICE same as mine tutorial answer thanks =)

Answer 12

you're welcome! So you get it?

Answer 13

Then just back substitute for u = sinx

Answer 14

yea got the idea on it now working it out on my paper =)

Answer 15

cool!