What is the integral from 0 to pi of sqrt(θ^4+4θ^2) ?

The integral from 0 to pi of sqrt(θ^4+4θ^2) is ((pi^2+4)^{3/2}-8)/3

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integral from 0 to pi of sqrt(θ^4+4θ^2)

\int_{0}^{π} θ^{4} + 4 θ^{2} d θ

\frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

Decimal

14.55102 \dots

Solution steps

\int_{0}^{π} θ^{4} + 4 θ^{2} d θ

Multiply by the conjugate of

θ^{4} + 4 θ^{2} : \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}}

= \int_{0}^{π} \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}} d θ

Find undefined (singularity) points:

θ = 2

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{2} \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}} d θ + \int_{2}^{π} \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}} d θ

\int_{0}^{2} \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}} d θ = \frac{16 2 - 8}{3}

\int_{2}^{π} \frac{θ ^{8} - 16 θ ^{4}}{θ ^{4} - 4 θ ^{2}} d θ = \frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3}

= \frac{16 2 - 8}{3} + \frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3}

Simplify

\frac{16 2 - 8}{3} + \frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 16 2}{3} : \frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

= \frac{( π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

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What is the integral from 0 to pi of sqrt(θ^4+4θ^2) ?
The integral from 0 to pi of sqrt(θ^4+4θ^2) is ((pi^2+4)^{3/2}-8)/3