What is the integral from 0 to 1 of arccot(x) ?

The integral from 0 to 1 of arccot(x) is pi/4+1/2 ln(2)

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integral from 0 to 1 of arccot(x)

\int_{0}^{1} arccot (x) d x

\frac{π}{4} + \frac{1}{2} ln (2)

Decimal

1.13197 \dots

Solution steps

\int_{0}^{1} arccot (x) d x

Apply Integration By Parts

: [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{0}^{1}

= [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{0}^{1}

\int - \frac{x}{x ^{2} + 1} d x = - \frac{1}{2} ln x^{2} + 1

= [x arccot (x) - (- \frac{1}{2} ln x^{2} + 1)]_{0}^{1}

Simplify

= [x arccot (x) + \frac{1}{2} ln x^{2} + 1]_{0}^{1}

Compute the boundaries:

\frac{π}{4} + \frac{1}{2} ln (2)

= \frac{π}{4} + \frac{1}{2} ln (2)

\int_{1}^{4} x x - 1 d x

\int_{0}^{x} 1 - t^{2} d t

\int_{0}^{2} 2 x + x^{2} - x^{3} d x

\int_{0}^{5} ∣ 1 - x ∣ d x

\int_{0}^{2} (t - 1) e^{(t - 1)^{2}} d t

What is the integral from 0 to 1 of arccot(x) ?
The integral from 0 to 1 of arccot(x) is pi/4+1/2 ln(2)