What is the integral from 0 to pi/2 of sin(x+9) ?

The integral from 0 to pi/2 of sin(x+9) is -cos(pi/2+9)+cos(9)

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integral from 0 to pi/2 of sin(x+9)

\int_{0}^{\frac{π}{2}} sin (x + 9) d x

- cos (\frac{π}{2} + 9) + cos (9)

Decimal Notation

- 0.49901 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin (x + 9) d x

Apply u-substitution

: \int_{9}^{\frac{π}{2} + 9} sin (u) d u

= \int_{9}^{\frac{π}{2} + 9} sin (u) d u

Use the common integral:

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

= [- cos (u)]_{9}^{\frac{π}{2} + 9}

Compute the boundaries:

- cos (\frac{π}{2} + 9) + cos (9)

= - cos (\frac{π}{2} + 9) + cos (9)

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\int_{0}^{1} x^{4} sin (π x) d x

What is the integral from 0 to pi/2 of sin(x+9) ?
The integral from 0 to pi/2 of sin(x+9) is -cos(pi/2+9)+cos(9)