Trig integration (cos(x)+sin(x)) / sin(2x)

In summary, the conversation discusses the solving of a problem involving the integral of a trigonometric expression. The participant initially uses a trigonometric property to simplify the expression, but later realizes that they may have made a mistake. Another participant points out the mistake and offers a correction. The first participant then graphs the original and corrected expressions and concludes that they are equivalent. The conversation also briefly touches on the different forms of the integral of csc(x).
  • #1
SpicyPepper
20
0
Prepping for my test, and I can't seem to solve this problem.

Homework Statement



[tex]\int \frac{cos(x) + sin(x)}{sin(2x)}dx[/tex]

Homework Equations



Not sure if it led me astray, but I used the trig. property:

[tex]sin(2x) = 2sin(x)cos(x)[/tex]

The Attempt at a Solution



[tex]= \int \frac{cos(x) + sin(x)}{2sin(x)cos(x)}dx[/tex]

[tex]= 1/2 \int(csc(x) + sec(x))dx[/tex]

[tex]= 1/2ln|csc(x) - cot(x)| + 1/2ln|sec(x) + tan(x)| + C[/tex]

[tex]= 1/2 ln|csc(x)sec(x) + sec(x) - csc(x) - 1| + C[/tex]

I feel like this is wrong, but not sure where I'm messing up. Maybe my trig. property isn't the way to go.
 
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  • #2
Looks good to me except you need + C.:cool:
 
  • #3
I would keep the answer in the form of the line before the last one. There's one little mistake in that line, though.
∫cscx dx = -ln|cscx + cotx| + C, not ln|cscx - cotx|
 
  • #4
Bohrok said:
I would keep the answer in the form of the line before the last one. There's one little mistake in that line, though.
∫cscx dx = -ln|cscx + cotx| + C, not ln|cscx - cotx|

Uhh...

[tex]LHS=ln\left(\frac{1}{cscx+cotx}\right)[/tex]

[tex]=ln\left(\frac{1}{\frac{1}{sinx}+\frac{cosx}{sinx}}\right)[/tex]

[tex]=ln\left(\frac{sinx}{1+cosx}\right)[/tex]


[tex]RHS=ln\left(\frac{1}{sinx}-\frac{cosx}{sinx}\right)[/tex]

[tex]=ln\left(\frac{1-cosx}{sinx}\right)[/tex]


Now, let's assume:[tex]ln\left(\frac{sinx}{1+cosx}\right)=ln\left(\frac{1-cosx}{sinx}\right)[/tex]

[tex]ln\left(\frac{sinx}{1+cosx}\right)-ln\left(\frac{1-cosx}{sinx}\right)=0[/tex]

[tex]ln\left(\frac{(\frac{sinx}{1+cosx})}{(\frac{1-cosx}{sinx})}\right)=0[/tex]

[tex]ln\left(\frac{sin^2x}{1-cos^2x}\right)=0[/tex]

[tex]ln\left(\frac{1-cos^2x}{1-cos^2x}\right)=0[/tex]

[tex]ln(1)=0[/tex]

Therefore, they are equivalent.
 
  • #5
Graphically, it looks like they are the same if you ignore the absolute values. LHS - RHS looks like 0 on wolframalpha, but with absolute values, the graph of their difference is not 0.
 
  • #6
I re-checked my work and didn't seem to find any errors or possibilities of extraneous solutions etc. so I graphed it as well, but on graphmatica. It seems that with or without absolute values, both functions are equal. If you believe you haven't made a mistake, please post a picture of what you see on wolfram :smile:
 
  • #8
Ok, there is no difference between our integrals of cscx. I didn't recognize the other form immediately (I never worked with ∫cscx dx by itself either) so I thought it was wrong. Looking at the table of integrals in my calc book, I see that same integral, ln|cscx - cotx| and I did find it by looking on the internet, but apparently that form is not as common as the other one.

I learn something new almost every day on this site. :smile:
 
  • #9
cool discussion guys. I just worked the integral of csc(x) on my own, and I came up with that funky version. Didn't realize it was less common :).
 

What is trig integration?

Trig integration is a mathematical process of finding the integral of a trigonometric function. It involves finding the antiderivative or indefinite integral of the given function.

What is the formula for integrating (cos(x)+sin(x)) / sin(2x)?

The formula for integrating (cos(x)+sin(x)) / sin(2x) is ∫ (cos(x)+sin(x)) / sin(2x) dx = -1/2 ln(csc(x) - cot(x)) + C, where C is the constant of integration.

What is the technique for integrating (cos(x)+sin(x)) / sin(2x)?

The technique for integrating (cos(x)+sin(x)) / sin(2x) is to use the substitution method. Let u = sin(x), then du = cos(x) dx. The integral then becomes ∫ (du + u) / sin(2x) = ∫ (1 + u) / sin(2x) dx, which can be solved using the formula mentioned above.

What are the common mistakes made when integrating (cos(x)+sin(x)) / sin(2x)?

One common mistake is forgetting to apply the chain rule when using the substitution method. It is also important to remember to add the constant of integration at the end. Another mistake is incorrectly applying the trigonometric identities, which can lead to incorrect results.

What are the real-life applications of trig integration?

Trig integration has many real-life applications in fields such as physics, engineering, and economics. It can be used to calculate the area under a curve, which is useful in determining volumes, work, and average values. It is also used in the study of periodic phenomena such as waves, vibrations, and electrical signals.

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