What is the integral from 0 to 2pi of sqrt(1+t^2) ?

The integral from 0 to 2pi of sqrt(1+t^2) is pisqrt(4pi^2+1)+1/2 ln(2pi+sqrt(4pi^2+1))

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integral from 0 to 2pi of sqrt(1+t^2)

\int_{0}^{2 π} 1 + t^{2} d t

π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Decimal Notation

21.25629 \dots

Solution steps

\int_{0}^{2 π} 1 + t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2 π)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2 π)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2 π)} sec (u) d u

\int_{0}^{a r c t a n (2 π)} sec (u) d u = ln (2 π + 4 π^{2} + 1)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Compute the boundaries:

π 4 π^{2} + 1

= π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

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\int_{0}^{1} \frac{2}{x ^{3}} d x

\int_{0}^{1} - \frac{( x - 1 ) ^{3}}{3} d x

\int_{0}^{1} 2 π (2 - x) (x - x^{2}) d x

What is the integral from 0 to 2pi of sqrt(1+t^2) ?
The integral from 0 to 2pi of sqrt(1+t^2) is pisqrt(4pi^2+1)+1/2 ln(2pi+sqrt(4pi^2+1))