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Indefinite Trig Integral

  • Thread starter planauts
  • Start date
  • #1
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Homework Statement



[tex]\int \frac{cos x}{sin^2(x) - sin(x)-6}[/tex]


2. The attempt at a solution
I first tried factoring the denominator.
[tex]\int \frac{cos x}{(sin(x) -3)(sin(x)+2)}[/tex]

The first thing that came to my mind was Partial Factoring but I don't think it would work in this case.


Thanks in advance!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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I would start by substituting u=sin(x) and then worry about the partial fractions in the variable u.
 
Last edited:
  • #3
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Let:
u = sin(x)
du = cos(x) dx

[tex]\int \frac{du}{(u-3)(u+2)}[/tex]

After doing Partial Fractions, I ended up with:
[tex]\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)[/tex]

Thanks for your help. I think that is the correct answer.
 
  • #4
gb7nash
Homework Helper
805
1
Let:
u = sin(x)
du = cos(x) dx

[tex]\int \frac{du}{(u-3)(u+2)}[/tex]

After doing Partial Fractions, I ended up with:
[tex]\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)[/tex]

Thanks for your help. I think that is the correct answer.
Also note you can factor out a 1/5 and you get:

[tex]\frac{1}{5} ln(\frac{sinx-3}{sinx+2})[/tex]
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
Let:
u = sin(x)
du = cos(x) dx

[tex]\int \frac{du}{(u-3)(u+2)}[/tex]

After doing Partial Fractions, I ended up with:
[tex]\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)[/tex]

Thanks for your help. I think that is the correct answer.
I'm really sure it's right.
 
  • #6
Char. Limit
Gold Member
1,204
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Don't forget the +C!
 
  • #7
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Don't forget the +C!
Yes you are right, I would have lost 1/2 a mark on a test.

[tex]\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2) + C[/tex]

OR as gb7nash pointed out
[tex]
\frac{1}{5} ln(\frac{sinx-3}{sinx+2} + C)
[/tex]

Thanks everyone!
 

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