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

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

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integral from 0 to pi/2 of (sin(t))

\int_{0}^{\frac{π}{2}} (sin (t)) d t

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (t) d t

Use the common integral:

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

= [- cos (t)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= 1

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

\int_{0}^{1} x^{2} 6 - 2 x^{3} d x

\int_{0}^{y} 2 x yz d z

\int_{0}^{1} e^{- i wt} d t

\int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{4}{5}}} d x

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