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I don't know latex but I have my work written out in this image. The answer seems to be incorrect and I am aware of the trig substitution trick that is usually used to solve this integral. I'm not interested in memorizing such a substitution and I don't need the answer since I can look it up if I need to. I'm wondering where my work went wrong here, and how to fix it so that memorizing dubious substitutions won't be necessary.

I also attempted this integral by writing sin(x) in exponential form via eulers identity and ended up with ln(abs((e^ix +1)/(e^ix -1 ))) +C as my answer, but here I am stuck trying to convert this back into trig terms. I suspect this answer is incorrect as well, probably for the same reason my first answer is incorrect.

What are non-memorization based ways of solving this integral and why is my approach incorrect?

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up vote 1 down vote accepted

The form of answers may vary in indefinite integration. We can always apply derivative to validate the result.

As $\cos2y=2\cos^2y-1=1-2\sin^2y,$

$$\left|\frac{\cos x-1}{\cos x+1}\right|=\left|\frac{-2\sin^2 \frac x2}{2\cos^ 2\frac x2}\right|=|-1|\tan^2\frac x2=\tan^2\frac x2$$

$$\text{So,} \ln\left|\frac{\cos x-1}{\cos x+1}\right|=\ln\left|\tan^2\frac x2\right|=2\ln \left|\tan\frac x2\right|$$

Alternatively, you can use Weierstrass substitution

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Thank you, that was a very clear explanation. I've never seen that last substitution before. It's very amazing, how does one come up with such a parameterization of the unit circle? – R R Apr 2 '13 at 15:20
@RR, not sure if I understood your query correctly. If you mean $x^2+y^2=1$ we can parametrize with $\cos\theta,\sin\theta$ – lab bhattacharjee Apr 2 '13 at 15:22
@RR, may also have a look at – lab bhattacharjee Apr 2 '13 at 15:23
I meant on the last section of the Wiesterstrass substitution article where it talks about how this change of variables captures (mostly) all of the points on the unit circle. Perhaps I just need more time to read through it. – R R Apr 2 '13 at 15:28

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