What is the integral from 0 to cot(x) of 1/(1+x^2) ?

The integral from 0 to cot(x) of 1/(1+x^2) is arctan(cot(x))

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integral from 0 to cot(x) of 1/(1+x^2)

\int_{0}^{c o t (x)} \frac{1}{1 + x ^{2}} d x

arctan (cot (x))

Solution steps

\int_{0}^{c o t (x)} \frac{1}{1 + x ^{2}} d x

Use the common integral:

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

= [arctan (x)]_{0}^{c o t (x)}

Compute the boundaries:

arctan (cot (x))

= arctan (cot (x))

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\int_{5}^{6} 3 t^{2} - 24 t + 45 d t

What is the integral from 0 to cot(x) of 1/(1+x^2) ?
The integral from 0 to cot(x) of 1/(1+x^2) is arctan(cot(x))