What is the integral from 0 to pi/2 of sin(sqrt(2y)) ?

The integral from 0 to pi/2 of sin(sqrt(2y)) is sin(sqrt(pi))-sqrt(pi)cos(sqrt(pi))

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integral from 0 to pi/2 of sin(sqrt(2y))

\int_{0}^{\frac{π}{2}} sin (2 y) d y

sin (π) - π cos (π)

Decimal

1.33474 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 y) d y

Apply u-substitution

: \int_{0}^{π} u sin (u) d u

= \int_{0}^{π} u sin (u) d u

Apply Integration By Parts

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

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

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

= [- u cos (u) - (- sin (u))]_{0}^{π}

Simplify

= [- u cos (u) + sin (u)]_{0}^{π}

Compute the boundaries:

sin (π) - π cos (π)

= sin (π) - π cos (π)

\int_{\frac{1}{3}}^{3} 6 - \frac{6}{x ^{2}} d x

\int_{- π}^{0} - sin (x) cos (n x) d x

\int_{0}^{1} 6 x d x

\int_{0}^{t^{2}} u d u

\int_{2}^{3} \frac{( 2 x - 3 )}{4 x - x ^{2}} d x

What is the integral from 0 to pi/2 of sin(sqrt(2y)) ?
The integral from 0 to pi/2 of sin(sqrt(2y)) is sin(sqrt(pi))-sqrt(pi)cos(sqrt(pi))