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mcastillo356

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- TL;DR Summary
- Natural logarithm is always the answer, but I would like an explanation of the different answers.

Hi, PF

"The method of substitution is often useful for evaluating trigonometric integrals" (Calculus, R. Adams and Christopher Essex, 7th ed)

$$\displaystyle\int{\csc{(x)}}=\displaystyle\int{\displaystyle\frac{1}{\sin{(x)}}}=-\ln{(|\csc{(x)}+\cot{(x)}|)}=\ln{(|\csc{(x)}-\cot{(x)}|)}=\ln{\Bigg|\tan\Bigg(\displaystyle\frac{1}{2}x\Bigg)\Bigg|}$$

The textbook recommends to memorize all of this: should I, or give a try and outline a proof? I mean, this is the first time I'm suggested to memorize. But, just some questions:

(i) Why the negative natural logarithm of the sum inside the absolute value is equal to the positive one, provided the subtraction?.

(ii) Could I approach somehow to some kind of explanation about the evaluation of this integral?.

Greetings!

PD: Post without preview.

**Trigonometric Integrals**"The method of substitution is often useful for evaluating trigonometric integrals" (Calculus, R. Adams and Christopher Essex, 7th ed)

**Integral of cosecant**$$\displaystyle\int{\csc{(x)}}=\displaystyle\int{\displaystyle\frac{1}{\sin{(x)}}}=-\ln{(|\csc{(x)}+\cot{(x)}|)}=\ln{(|\csc{(x)}-\cot{(x)}|)}=\ln{\Bigg|\tan\Bigg(\displaystyle\frac{1}{2}x\Bigg)\Bigg|}$$

The textbook recommends to memorize all of this: should I, or give a try and outline a proof? I mean, this is the first time I'm suggested to memorize. But, just some questions:

(i) Why the negative natural logarithm of the sum inside the absolute value is equal to the positive one, provided the subtraction?.

(ii) Could I approach somehow to some kind of explanation about the evaluation of this integral?.

Greetings!

PD: Post without preview.