What is the integral from 0 to 1 of sqrt(5+4t^2) ?

The integral from 0 to 1 of sqrt(5+4t^2) is 1/8 (12+5ln(5))

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integral from 0 to 1 of sqrt(5+4t^2)

\int_{0}^{1} 5 + 4 t^{2} d t

\frac{1}{8} (12 + 5 ln (5))

Decimal

2.50589 \dots

Solution steps

\int_{0}^{1} 5 + 4 t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{2}{5})} \frac{5}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{2}{5})} \frac{5}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{5}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{5})} sec^{3} (u) d u

Apply Integral Reduction

= \frac{5}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{5})} sec (u) d u)

\int_{0}^{a r c t a n (\frac{2}{5})} sec (u) d u = \frac{1}{2} ln (5)

= \frac{5}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + \frac{1}{2} \cdot \frac{1}{2} ln (5))

Simplify

\frac{5}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + \frac{1}{2} \cdot \frac{1}{2} ln (5)) : \frac{1}{8} (20 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + 5 ln (5))

= \frac{1}{8} (20 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + 5 ln (5))

Compute the boundaries:

\frac{3}{5}

= \frac{1}{8} (20 \cdot \frac{3}{5} + 5 ln (5))

Simplify

= \frac{1}{8} (12 + 5 ln (5))

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What is the integral from 0 to 1 of sqrt(5+4t^2) ?
The integral from 0 to 1 of sqrt(5+4t^2) is 1/8 (12+5ln(5))