What is the integral from 0 to pi/2 of sin(x^{1/2}) ?

The integral from 0 to pi/2 of sin(x^{1/2}) is 2(sin((pi/2)^{1/2})-(pi/2)^{1/2}cos((pi/2)^{1/2}))

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integral from 0 to pi/2 of sin(x^{1/2})

\int_{0}^{\frac{π}{2}} sin (x^{\frac{1}{2}}) d x

2 (sin ((\frac{π}{2})^{\frac{1}{2}}) - (\frac{π}{2})^{\frac{1}{2}} cos ((\frac{π}{2})^{\frac{1}{2}}))

Decimal Notation

1.11754 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin (x^{\frac{1}{2}}) d x

Apply u-substitution

: \int_{0}^{(\frac{π}{2})^{\frac{1}{2}}} sin (u) \cdot 2 u d u

= \int_{0}^{(\frac{π}{2})^{\frac{1}{2}}} 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}^{(\frac{π}{2})^{\frac{1}{2}}} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{0}^{(\frac{π}{2})^{\frac{1}{2}}}

= 2 [- u cos (u) - \int - cos (u) d u]_{0}^{(\frac{π}{2})^{\frac{1}{2}}}

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

= 2 [- u cos (u) - (- sin (u))]_{0}^{(\frac{π}{2})^{\frac{1}{2}}}

Simplify

= 2 [- u cos (u) + sin (u)]_{0}^{(\frac{π}{2})^{\frac{1}{2}}}

Compute the boundaries:

sin ((\frac{π}{2})^{\frac{1}{2}}) - (\frac{π}{2})^{\frac{1}{2}} cos ((\frac{π}{2})^{\frac{1}{2}})

= 2 (sin ((\frac{π}{2})^{\frac{1}{2}}) - (\frac{π}{2})^{\frac{1}{2}} cos ((\frac{π}{2})^{\frac{1}{2}}))

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What is the integral from 0 to pi/2 of sin(x^{1/2}) ?
The integral from 0 to pi/2 of sin(x^{1/2}) is 2(sin((pi/2)^{1/2})-(pi/2)^{1/2}cos((pi/2)^{1/2}))