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Trig anti-dervative

  • Thread starter jwxie
  • Start date
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1. Homework Statement

Int of sin x over - sin ^2 x of dx


3. The Attempt at a Solution

I don;t know if I have the right question. But I just couldn't reproduce the problem (from my answer)!!
please take a look at my note, see if you can reproduce the original question

i am talking about question #7
 

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Is this your integral?
[tex]\int \frac{sin x}{-sin^2 x}dx[/tex]

If so, this is the same as
[tex]\int -csc (x) dx[/tex]
If I recall correctly, this can be done using integration by parts.
 
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yeah i think that;s what it is
but if we use -csc x can it be done easily?
i mean cscx does not have any Anti-derv...
DSC04361.jpg
 
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But -sin^2(x) isn't equal to cos^2(x), so your first step is incorrect.
 
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then shouldn't it be int of sinx times -sin^2 x?
 
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then shouldn't it be int of sinx times -sin^2 x?
I don't know what you mean. This is what you wrote in your first post in this thread:
Int of sin x over - sin ^2 x of dx
By "over" I assume you mean the quotient of sin(x) and -sin^2(x), which is what I showed in the integral.
 
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i am sorry, i am referring to your #4
you said my first step was wrong

now, i knew the mistake, and this is what i did

original question:
Int of sin x over - sin ^2 x of dx

first step, change the bottom, -sin^2 (x) to this form --> 1 over csc^2 (x)

so the entire int will become

int of sin (x) times csc^2 (x)

because 1/sinx = cscx, then 1 / -sin^2 (x) = csc^2 (x)
am i correct?
 
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i am sorry, i am referring to your #4
you said my first step was wrong

now, i knew the mistake, and this is what i did

original question:
Int of sin x over - sin ^2 x of dx
Or, using inline LaTeX tags,
[itex]\int sin(x)/(-sin^2(x)) dx[/itex]
first step, change the bottom, -sin^2 (x) to this form --> 1 over csc^2 (x)
You've lost a sign. -sin^2(x) = -1/csc^2(x)
so the entire int will become

int of sin (x) times csc^2 (x)
Or [itex]\int sin(x)(-csc^2(x)) dx[/itex]
because 1/sinx = cscx, then 1 / -sin^2 (x) = csc^2 (x)
am i correct?
No.
1 / -sin^2 (x) = -csc^2 (x)
 
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You're going around in circles. The integrand simplifies to -1/sin x or -csc x.

It is not at all obvious how to integrate csc x. Try multiplying the integrand by a certain factor which allows you to make a substitution, but which doesn't change the value of the integrand. Hint: the final result involves a logarithm.
 

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