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

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

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integral from 0 to pi/2 of sin(2θ)

\int_{0}^{\frac{π}{2}} sin (2 θ) d θ

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 θ) d θ

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

\frac{d}{d x} (3 ln (sec (x) + tan (x)))

y^{^{'}} - e^{x + 1} = 0

\int sec (\frac{x}{4}) d x

y^{^{''}} + 9 y = 2 sin (x) + cos (x)

\frac{d y}{d x} - x y = x

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