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

The integral from 0 to pi^2 of sin(sqrt(x)) is 2pi

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

\int_{0}^{π^{2}} sin (x) d x

2 π

Decimal

6.28318 \dots

Solution steps

\int_{0}^{π^{2}} sin (x) d x

Apply u-substitution

= \int_{0}^{π} sin (u) \cdot 2 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}^{π} sin (u) u d u

Apply Integration By Parts

= 2 [- u cos (u) - \int - cos (u) d u]_{0}^{π}

\int - cos (u) d u = - sin (u)

= 2 [- u cos (u) - (- sin (u))]_{0}^{π}

Simplify

= 2 [- u cos (u) + sin (u)]_{0}^{π}

Compute the boundaries:

π

= 2 π

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