Trig integration question

[lilshai]

Hello:

I am trying to integrate dx/(sin²x) and so I used a trig identity to simplify matters (sin²x = (1-cos 2x)/2. So I pull the 2 outside of the integral and I have dx/(1-cos 2x) left to integrate. Any tips on how to proceed from here on? Thanks.

 

[Gene]

That would be a good approach if it were in the numerator, but since it is in the denominator
1/sin²(x) = csc²(x)
and
d(cot(x))= - csc²(x)dx

 

[lilshai]

Yea, I guess that works if you know what the derivative of cotangent is. I was trying to do it in terms of sine's and cosine's. I was actually reading somewhere else how you can integrate 1/sinx as follows:

dx/sinx = sinx dx/sinx²
: = sinx dx/(1-cos²x)

:now you do "u substitution" - u=cosx, du=-sinx dx
:
:Then you end up with -du/(1-u²)

Any ideas on how to integrate that?

 

[Gene]

Since you are integrating you don't have to know what the derivitive is. The anti derivitive of the derivitive is the function. The answer is
integral(dx/sin(x)^2) = -cot(x)+C.
If you want to do it with sine and cosines
d(u/v)=(vdu-udv)/v^2
with u=-cos(x) and v=sin(x) will do it.

Part two:
You end up with a negative inverse hyperbolic tangent with that integral. I've managed to avoid hypebolics almost completely. Is that another problem? I don't see a connection between integrating dx/sin(x)^2 and dx/sin(x)

 

[lilshai]

What I meant is, what if you didn't know that the antiderivative of csc²x was -cot(x)? Then how would you do it? That's why I was trying to show the connection of the other approach, because they introduced sin x so that they could deal with just sine's and cosine's.

 

[Gene]

Gotcha.
All I can say is you have to think about all the derivitives you know and try to pick one that comes close to fitting the problem. The square in the denominator might bring to mind d(u/v) 'cause that has a the square you are looking for. If you are very clever you might replace the 1 by sin^2+cos^2. Then you would have to recognize the u*dv and v*du hiding in the numerator. You know v=sin(x) from the sin(x)^2 so dv=cos(x). That brings you to (u*cos(x)-sin(x)du)/sin(x)^2. To get the cos^2, u=cos(x). Then you are home free.
That's the best rationalization I can come up with.
Can any other tutor do better?
---------------------
Gene

 

[lilshai]

Isn't that the quotient rule for derivatives? I thought we had to use integration by parts. Letting the 1 = sin²x + cos²x is a great idea, but wouldn't I still have to integrate [sin²x + cos²x]/sin²x ?

 

[Gene]

Integration is just the reverse of differentiation. Once you can write
d(u/v)=f(u,v)
the integral of f(u,v) is u/v.
That's what we were doing with the magic we did to 1/sin(x)^2 to get it to be
(sin(x)d(cos(x) -cos(x)d(sin(x))/sin(x)^2 = (vdu-udv)/v^2 =
-d(u/v) =
-d(cos(x)/sin(x)) =
-cot(x)

 

[lilshai]

(vdu-udv)/v^2 =
-d(u/v) =
-d(cos(x)/sin(x)) =
-cot(x)

Hello, I am almost there. I see what you are saying now. Can you explain how you got those last few steps? I am just left with -sin²x/sin²x - cos²x/sin²x = -1-cot²x. What am I doing wrong?

 

[Gene]

sin = -d(cos)
cos = d(sin)

Starting at the beginning, leaving the sin(x)^2 in the denominator alone, just doing the numerator:

1 =
cos² + sin² =
cos*cos + sin*sin
Substituting those first two identities for the second of each of the two terms gives
cos*(d(sin)) + sin*(-d(cos)) =
cos*d(sin) - sin*d(cos) =
-(sin*d(cos)-cos*d(sin))
Except for the minus in front that is now in the form of
v*d(u) - u*d(v)
which is where we want to get for the numerator of
d(u/v) = (v*d(u) - u*d(v))/v²
where u=cos & v=sin. It still = 1/sin² so we have
-d(cos/sin)=1/sin²
integrating both sides gives
-cos/sin=integral(1/sin²)
Which is what we wanted to find. Finally, if you want,
-cos/sin = -cot

 

[lilshai]

Hi Gene,
ohhhh ok, I finally get it now! I completely neglected the d(u/v) part on the left hand side. Thanks for taking the time to type everything out and being so patient, I appreciate it. That was a very clever trick, the 1 = sin²+cos² :wink: