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

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

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

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

π - 2

Solution steps

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

Apply Integration By Parts

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

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

\int - 2 y cos (y) d y = - 2 (y sin (y) + cos (y))

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

Simplify

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

Compute the boundaries:

π - 2

= π - 2

\int_{0}^{x^{2}} ln (1 + t) d t

\int_{- 2}^{0} \frac{1}{e ^{- 2 x}} d x

\int_{0}^{1} e^{x} + 1 d x

\int_{- 2}^{1} (x^{2} + x - 2) d x

\int_{0}^{2 π} 2 d x

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