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

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

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

\int_{0}^{π} 1 + x^{2} d x

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

Decimal

6.10991 \dots

Solution steps

\int_{0}^{π} 1 + x^{2} d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π π ^{2} + 1}{2}

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

Simplify

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

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