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

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

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

\int_{0}^{\frac{π}{2}} sin (2 x) d x

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 x) d x

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

3 t^{4} - 2 t^{2} + 1

cos (2 t)

y = 16 - x^{2}

\frac{\partial}{\partial x} ((a - x)^{2} + b (y - x^{2})^{2})

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