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

The integral from 0 to pi/2 of sin(φ) is 1

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integral from 0 to pi/2 of sin(φ)

\int_{0}^{\frac{π}{2}} sin (φ) d φ

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (φ) d φ

Use the common integral:

\int sin (φ) d φ = - cos (φ)

= [- cos (φ)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 1

\int_{0}^{1} (x^{2}) - (x) d x

\int_{- 3}^{2} (12 - x^{2} - x) d x

\int_{1}^{2.5} x d x

\int_{x^{2}}^{x} x y d y

\int_{0}^{4} x \frac{3}{16} x d x

What is the integral from 0 to pi/2 of sin(φ) ?
The integral from 0 to pi/2 of sin(φ) is 1