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

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

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

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

- \frac{( 2 - ln ( - 1 + 2 ) ) a ^{2}}{2}

Solution steps

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

Apply Trigonometric Substitution

= \int_{0}^{- \frac{π}{4}} a^{2} sec^{3} (u) d u

Take the constant out:

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

= a^{2} \cdot \int_{0}^{- \frac{π}{4}} sec^{3} (u) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= a^{2} (- \int_{- \frac{π}{4}}^{0} sec^{3} (u) d u)

Apply Integral Reduction

= a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{0} sec (u) d u))

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0}))

Simplify

a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})) : - a^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})

= - a^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})

Compute the boundaries:

\frac{1}{2}

Compute the boundaries:

- ln (2 - 1)

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

Simplify

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

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