What is the integral from 0 to 1 of (sin(pix))^2 ?

The integral from 0 to 1 of (sin(pix))^2 is 1/2

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integral from 0 to 1 of (sin(pix))^2

\int_{0}^{1} (sin (π x))^{2} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{1} (sin (π x))^{2} d x

Simplify

= \int_{0}^{1} sin^{2} (π x) d x

Rewrite using trig identities

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

Take the constant out:

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

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

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

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} cos (2 π x) d x = 0

= \frac{1}{2} (1 - 0)

Simplify

= \frac{1}{2}

\int (9 x^{2} - 4 x) e^{2 x} d x

\int e^{- 4 x} x d x

\int \frac{2 x + 3}{2 x + 1} d x

\frac{\partial}{\partial y} (e^{x + y})

x \to 2 lim (9 - x^{2})

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