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Q. 14

Expert-verifiedFound in: Page 362

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Verify that$\int \mathrm{cot}xdx=\mathrm{ln}\left(\mathrm{sin}x\right)+C$. (Do not try to solve the integral from scratch.)

It is verified that $\int \mathrm{cot}xdx=\mathrm{ln}\left(\mathrm{sin}x\right)+C.$

The given integral is $\int \mathrm{cot}xdx=\mathrm{ln}\left(\mathrm{sin}x\right)+C.$

We can write the differentiation as,

$\frac{d}{dx}\mathrm{ln}\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}x}\left(\mathrm{cos}x\right)$

$=\frac{\mathrm{cos}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=cotx$

Hence, $\mathrm{ln}\left(\mathrm{sin}x\right)\text{is an anti-derivative of}\mathrm{cot}x\text{.}$

Therefore,

$\int \mathrm{cot}xdx=\mathrm{ln}\left(\mathrm{sin}x\right)+C$

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