trigonometric integral

[ericfromcowtown]

integral cos(x)/((sin(x)^2)-2sin(x-8)) dx. I'm not sure where to begin with this one. Substituting u=x-8 just transfers the complexity from one term to another. I'd like to substitute u=sinx, so that du=cosx dx, but how do I then deal with the 2sin(x-8). I can't think of any trig identities that will help me initially. Help, I'm stuck. Can anyone point me in the right direction? Thanks.

 

[srmichael]

integral cos(x)/((sin(x)^2)-2sin(x-8)) dx. I'm not sure where to begin with this one. Substituting u=x-8 just transfers the complexity from one term to another. I'd like to substitute u=sinx, so that du=cosx dx, but how do I then deal with the 2sin(x-8). I can't think of any trig identities that will help me initially. Help, I'm stuck. Can anyone point me in the right direction? Thanks.

You sure the denominator isn't \(\displaystyle \sin^2(x)-2\sin(x)-8\)?? Not that that makes the integral any easier, but it is a little more realistic.

 

[ericfromcowtown]

You sure the denominator isn't \(\displaystyle \sin^2(x)-2\sin(x)-8\)?? Not that that makes the integral any easier, but it is a little more realistic.

Yes, it is! Sorry. As you said, that might not make the integral easier, but at least I know where to start. I'll have to work on this one later today and see where substituting u=sinx, du=cosx dx takes me.

 

[Deleted member 4993]

You sure the denominator isn't \(\displaystyle \sin^2(x)-2\sin(x)-8\)?? Not that that makes the integral any easier, but it is a little more realistic.

I think that does make it lot easier:

\(\displaystyle \displaystyle \int \frac{cos(x)}{sin^2(x)-2sin(x)-8} dx\)


\(\displaystyle = \ \displaystyle \int \frac{cos(x)}{[sin(x)+2][sin(x)-4]} dx\)

\(\displaystyle = \ \displaystyle \frac{1}{6} \int \left [ \frac{cos(x)}{sin(x)-4} \ - \ \frac{cos(x)}{sin(x)+2}\right ] dx\)

and so on....

 

[ericfromcowtown]

I think that does make it lot easier:

\(\displaystyle \displaystyle \int \frac{cos(x)}{sin^2(x)-2sin(x)-8} dx\)


\(\displaystyle = \ \displaystyle \int \frac{cos(x)}{[sin(x)+2][sin(x)-4]} dx\)

\(\displaystyle = \ \displaystyle \frac{1}{6} \int \left [ \frac{cos(x)}{sin(x)-4} \ - \ \frac{cos(x)}{sin(x)+2}\right ] dx\)

and so on....

Thanks. I wouldn't have thought to use quadratic factoring like that when it comes to sin^2(x) in contrast to x^2, but it makes sense. I don't think I've seen that before.

At that point we're just left to integrate and should get 1/6 ln (sin(x)-4) - 1/6 ln (sin(x)+2) + C, correct?

I'm taking a "Calculus II" course via a less-than-stellar correspondence school and I'm relying largely on youtube and websites like this to remove the rust from my 40 year old brain and fill in where TA support is lacking.

 

[Deleted member 4993]

You indicated that, you wanted to substitute u = sin(x). You would have come to same expression after integration and back-substitution. You could take one more step to write:

\(\displaystyle \displaystyle \ = \ ln \left [C *\frac{sin(x)-4}{sin(x)+2}\right ]^{\dfrac{1}{6}}\)

I suggest this form because - as such sin(x) - 4 <0 → ln[sin(x) - 4] is undefined in real domain. In this case, we must have C < 0.

 

[HallsofIvy]

Thanks. I wouldn't have thought to use quadratic factoring like that when it comes to sin^2(x) in contrast to x^2, but it makes sense. I don't think I've seen that before.

Actually, the first thing you should have thought of, seeing that "cos(x)" in the numerator and "sin(x)" in the denominator, was the substitutioin u= sin(x). Then du= cos(x) and the integral becomes
\(\displaystyle \int\frac{du}{u^2- 2u- 8}\)
That should make the idea of factoring the denominator easier.

At that point we're just left to integrate and should get 1/6 ln (sin(x)-4) - 1/6 ln (sin(x)+2) + C, correct?

I'm taking a "Calculus II" course via a less-than-stellar correspondence school and I'm relying largely on youtube and websites like this to remove the rust from my 40 year old brain and fill in where TA support is lacking.

(If this were "sin(x- 8)", my first thought was "sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)" so that sin(x- 8)= cos(8)sin(x)- sin(8)cos(x).)

 

[stapel]

Actually, the first thing you should have thought of...was....

"Should" being a relative term, of course; there are a lot of things that are "obvious" after one has gained more experience! :lol:

 

[lookagain]

You indicated that, you wanted to substitute u = sin(x).


You would have come to same expression after integration and back-substitution. You could take one more step to write:

\(\displaystyle \displaystyle \ = \ ln \left [C *\frac{sin(x)-4}{sin(x)+2}\right ]^{\dfrac{1}{6}}\)

I suggest this form because - as such sin(x) - 4 <0 → ln[sin(x) - 4] is undefined in real domain. In this case, we must have C < 0.

But one of the purpose/roles of the absolute value bars is to be used in certain antiderivatives involving the use of the natural logarithm.
And I would free up the "+ C" to be written at the usual far right. And for style purposes, I would bring down the "1/6" to keep the
answer to fitting on two lines.


Instead, I would write something closer to \(\displaystyle \ \dfrac{1}{6}\ln\bigg|\dfrac{sin(x) - 4}{sin(x) + 2}\bigg| \ + \ C\)

But, as the value of the fraction inside the absolute value bars is necessarily negative, then just drop the absolute value bars,
and compensate with adjusting the fraction, such as with


\(\displaystyle \ \dfrac{1}{6}\ln\bigg[\dfrac{4 - sin(x)}{sin(x) + 2}\bigg] \ + \ C \)

or

\(\displaystyle \ \dfrac{1}{6}\ln\bigg(\dfrac{4 - sin(x)}{sin(x) + 2}\bigg) \ + \ C \)

 

[Deleted member 4993]

But one of the purpose/roles of the absolute value bars is to be used in certain antiderivatives involving the use of the natural logarithm.
And I would free up the "+ C" to be written at the usual far right. And for style purposes, I would bring down the "1/6" to keep the
answer to fitting on two lines.


Instead, I would write something closer to \(\displaystyle \ \dfrac{1}{6}\ln\bigg|\dfrac{sin(x) - 4}{sin(x) + 2}\bigg| \ + \ C\)

But, as the value of the fraction inside the absolute value bars is necessarily negative, then just drop the absolute value bars,
and compensate with adjusting the fraction, such as with


\(\displaystyle \ \dfrac{1}{6}\ln\bigg[\dfrac{4 - sin(x)}{sin(x) + 2}\bigg] \ + \ C \)

or

\(\displaystyle \ \dfrac{1}{6}\ln\bigg(\dfrac{4 - sin(x)}{sin(x) + 2}\bigg) \ + \ C \)

You could...

That is why it was a mere suggestion - not a command.

 

[lookagain]

You could...

That is why it was a mere suggestion - not a command.

That's overly clear. The above message in the quote box is unnecessary.
It's as if you are claiming that I was stating yours was other than a suggestion.

Another thing that I forgot to include is with the "+ C" version, there is no having to stipulate that C < 0

(when it is written as a multiplier on that fraction).