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## Homework Statement

Please find the anti-derivative of 1/sin(x) + cos(x) dx

## Homework Equations

csc(x)dx = -ln[csc(x) + cot(x)] + C

Thanks. Been at it for a while now.

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- Thread starter niravana21
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Inasmuch as you are studying calculus, it is to your benefit to learn how to express simple mathematical expressions clearly. In particular "1 over the sin and cos" is much less clear than "1 over the sum of the sine and cosine functions" or in symbols, 1/(sin(x) + cos(x)). The operation in the denominator is addition, not "and."f

- #1

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Please find the anti-derivative of 1/sin(x) + cos(x) dx

csc(x)dx = -ln[csc(x) + cot(x)] + C

Thanks. Been at it for a while now.

- #2

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- #4

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Sorry it was my mistake. Its 1 over the sin and cos

- #5

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Or in other words, the integrand is 1/(sin(x) + cos(x)).Sorry it was my mistake. Its 1 over the sin and cos

Parentheses are especially important when you're writing algebraic expressions on what is essentially a single line.

By writing 1/sin(x) + cos(x), most people would correctly interpret this as

[tex]\frac{1}{sin(x)} + cos(x)[/tex]

even though that's not what you intended.

- #6

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Or in other words, the integrand is 1/(sin(x) + cos(x)).

Parentheses are especially important when you're writing algebraic expressions on what is essentially a single line.

By writing 1/sin(x) + cos(x), most people would correctly interpret this as

[tex]\frac{1}{sin(x)} + cos(x)[/tex]

even though that's not what you intended.

yes i get it, but this has in no way helped me answer the question.

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yes i get it, but this has in no way helped me answer the question.

Did you miss my response in post 3?

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yes i get it, but this has in no way helped me answer the question.

But it made it harder for people on this forum to understand exactly what you were asking. The way you wrote it confused Cyosis, although Dick was able to translate what you wrote into what the problem actually was. My post was aimed at getting you to realize the importance of writing your problems so that people can easily understand them.

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Any ideas?

- #10

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But it made it harder for people on this forum to understand exactly what you were asking. The way you wrote it confused Cyosis, although Dick was able to translate what you wrote into what the problem actually was. My post was aimed at getting you to realize the importance of writing your problems so that people can easily understand them.

yes i realized that, so i quickly posted that it is 1 over the sin and cos. So do you have any ideas on solving that with the anti-deriv of csc(x)?

- #11

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So, putting y = pi/4 yields:

sin(x)+cos(x) = sqrt(2) sin(x+pi/4)

This means that you need to integrate 1/sin(x +pi/4) and you are already given the answer to that which you could also have derived very easily like this:

1/sin(x) = [sin^2(x/2) + cos^2(x/2)]/[2 sin(x/2)cos(x/2)] =

1/2 tan(x/2) + 1/2 cot(x/2) which is trivial to integrate.

- #12

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yes i realized that, so i quickly posted that it is 1 over the sin and cos.

Inasmuch as you are studying calculus, it is to your benefit to learn how to express simple mathematical expressions clearly. In particular "1 over the sin and cos" is much less clear than "1 over the sum of the sine and cosine functions" or in symbols, 1/(sin(x) + cos(x)). The operation in the denominator is addition, not "and."

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