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When you integrate $\int \cot \theta \csc^2 \theta d \theta$ it has two answers.

Using simple substitution you can solve like this:

$u=\cot \theta$, $du=-\csc^2 \theta d \theta$

$\int - u du=-\frac{u^2}{2}+C=-\frac{\cot^2 \theta}{2}+C$

However, if you apply trigonometical identities and express the integral like this: $\int \cot \theta \csc^2 \theta d \theta=\int \frac{\cos \theta}{\sin^3 \theta}d \theta$

You then can clearly use simple substitution to solve this but in a different way:

$u=\sin \theta$, $du=\cos \theta d \theta$

$\int \frac{du}{u^3}=-\frac{1}{2u^2}+C=-\frac{1}{2\sin^2 \theta}+C=-\frac{1}{2}\csc^2 \theta + C$

Now, when the integral is indefinite, both answers are exactly the same because of the trigonometrical identity: $\cot^2 \theta + 1 =\csc^2 \theta$. However, when the integral is definite from a point $a$ to a point $b$: $\int_a ^b \cot \theta \csc^2 \theta d \theta$. It gives two different answers.

So, what is the correct answer to the definite integral?

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    $\begingroup$ No, the two antiderivatives do not give two different answers to the definite integral. $\endgroup$ – David C. Ullrich Mar 31 '16 at 16:12
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    $\begingroup$ As David pointed out, the definite integral are the same. When you evaluate $\int_a^b f(x)dx=F(b)-F(a)$, if you replace $F$ with $F+c$, with $c$ constant, then the constant cancels. $\endgroup$ – bartgol Mar 31 '16 at 16:14
  • $\begingroup$ You are right... Fail $\endgroup$ – Gregorio Reyes Mar 31 '16 at 16:17

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