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

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

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

\int_{0}^{\frac{π}{2}} sin (2 t) d t

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 t) d t

Apply u-substitution

: \int_{0}^{π} sin (u) \frac{1}{2} d u

= \int_{0}^{π} sin (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{2} \cdot 2

Simplify

= 1

\int_{0}^{4} x + 1 d x

x \to 9 lim (3 - x)

\frac{d}{d x} (4 x^{3} - 7 x^{2})

y = 6 - 2 x, y = 2, x = 0

x \to 30 lim (\frac{x ^{2}}{6} - \frac{5}{x})

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