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

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

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

\int_{0}^{π} 2 + x^{2} d x

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

Decimal

6.95025 \dots

Solution steps

\int_{0}^{π} 2 + x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π π ^{2} + 2}{4}

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

Simplify

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

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