What is the integral from 0 to 8 of sin(sqrt(x)) ?

The integral from 0 to 8 of sin(sqrt(x)) is 2(sin(2sqrt(2))-2sqrt(2)cos(2sqrt(2)))

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integral from 0 to 8 of sin(sqrt(x))

\int_{0}^{8} sin (x) d x

2 (sin (22) - 22 cos (22))

Decimal Notation

5.99786 \dots

Solution steps

\int_{0}^{8} sin (x) d x

Apply u-substitution

: \int_{0}^{22} sin (u) \cdot 2 u d u

= \int_{0}^{22} 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}^{22} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{0}^{22}

= 2 [- u cos (u) - \int - cos (u) d u]_{0}^{22}

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

= 2 [- u cos (u) - (- sin (u))]_{0}^{22}

Simplify

= 2 [- u cos (u) + sin (u)]_{0}^{22}

Compute the boundaries:

sin (22) - 22 cos (22)

= 2 (sin (22) - 22 cos (22))

\int sin (3 x) cos (7 x) d x

x^{2} y^{^{''}} + 3 x y^{^{'}} - 24 y = 0, y (1) = 3, y^{^{'}} (1) = - 3

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x} + \frac{1}{y} + x y)

\int \frac{1}{1 + sin ( θ )} d θ

\int 2 x^{2} - x^{3} d x

What is the integral from 0 to 8 of sin(sqrt(x)) ?
The integral from 0 to 8 of sin(sqrt(x)) is 2(sin(2sqrt(2))-2sqrt(2)cos(2sqrt(2)))