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i need help to integrate $\frac{1}{\sin^2(x)} dx$. I tried integration by parts $u = \sin²x$ and $dv=1$ but still don't got an answer, thank you.

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$$\int\frac{dx}{\sin^2x}=\int\frac{\sec^2x}{\tan^2x}\ dx$$

Set $\tan x=u$

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  • $\begingroup$ Nice move(+1). Then spot it's a "chain rule backwards" as the derivative of $\tan x$ is $\sec^2 x$. $\endgroup$ Commented Dec 28, 2020 at 12:04
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What is the derivative of $\cot(x)$?

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