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

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

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

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

arctan (cot (x))

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (cot (x))

= arctan (cot (x))

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\int_{e}^{e^{2}} \frac{ln ( x ^{3} )}{x} d x

\int_{- 2}^{2} \frac{1}{x ^{3}} d x

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