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

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

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

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

- arctan (cot (x))

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- arctan (cot (x))

= - arctan (cot (x))

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\int_{0}^{π} 16 sin^{2} (x) d x

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