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

The integral from 0 to pi/2 of sin^2(2θ) is pi/4

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integral from 0 to pi/2 of sin^2(2θ)

\int_{0}^{\frac{π}{2}} sin^{2} (2 θ) d θ

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin^{2} (2 θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos (4 θ)) d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d θ - \int_{0}^{\frac{π}{2}} cos (4 θ) d θ)

\int_{0}^{\frac{π}{2}} 1 d θ = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 θ) d θ = 0

= \frac{1}{2} (\frac{π}{2} - 0)

Simplify

\frac{1}{2} (\frac{π}{2} - 0) : \frac{π}{4}

= \frac{π}{4}

\int_{0}^{\frac{1}{2}} x e^{3 x} d x

\int_{0}^{y} x^{2} e^{- x y} d x

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

\int_{\frac{π}{6}}^{2 π} cos (x) d x

\int_{2}^{5} x - 1 d x

What is the integral from 0 to pi/2 of sin^2(2θ) ?
The integral from 0 to pi/2 of sin^2(2θ) is pi/4