ahaanomegas
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Greetings! I hope this is the correct forum for my question (this is my first post, here).
The problem statement
Find: \displaystyle\int \cot x \ \mathrm{d}x
A Solution
We can re-write this integral into a convenient form: \displaystyle\int \cot x \ \mathrm{d}x = \displaystyle\int \dfrac{\cos x}{\sin x} \ \mathrm{d}x
Now, let u=\sin x. Then, du = \cos x \ \mathrm{d}x.
The integral becomes: \displaystyle\int \dfrac {du}{u} \ \mathrm{d}x and we integrate to get \ln |u| (ignoring the constant of integration, for now -- reason: simplicity).
We convert our result into a formula with respect to x:
My Question
If we use Integration by Parts, however, with u=\dfrac{1}{\sin x} and \mathrm{d}v = \cos x \ \mathrm{d}x and
Thanks in advance! :)
The problem statement
Find: \displaystyle\int \cot x \ \mathrm{d}x
A Solution
We can re-write this integral into a convenient form: \displaystyle\int \cot x \ \mathrm{d}x = \displaystyle\int \dfrac{\cos x}{\sin x} \ \mathrm{d}x
Now, let u=\sin x. Then, du = \cos x \ \mathrm{d}x.
The integral becomes: \displaystyle\int \dfrac {du}{u} \ \mathrm{d}x and we integrate to get \ln |u| (ignoring the constant of integration, for now -- reason: simplicity).
We convert our result into a formula with respect to x:
\boxed{\ln |\sin x| + \mathcal{C}}
My Question
If we use Integration by Parts, however, with u=\dfrac{1}{\sin x} and \mathrm{d}v = \cos x \ \mathrm{d}x and
\displaystyle\int u \ \mathrm{d}v = uv - \displaystyle\int v \ \mathrm{d}u,
we get 1=0. I can show my solution, if needed. Why is this? (A friend of mine pointed this out to me)Thanks in advance! :)