What is the integral from 0 to 3/4 of sin^2(pix) ?

The integral from 0 to 3/4 of sin^2(pix) is (3pi+2)/(8pi)

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integral from 0 to 3/4 of sin^2(pix)

\int_{0}^{\frac{3}{4}} sin^{2} (π x) d x

\frac{3 π + 2}{8 π}

Decimal Notation

0.45457 \dots

Solution steps

\int_{0}^{\frac{3}{4}} sin^{2} (π x) d x

Rewrite using trig identities

= \int_{0}^{\frac{3}{4}} \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}^{\frac{3}{4}} 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}^{\frac{3}{4}} 1 d x - \int_{0}^{\frac{3}{4}} cos (2 π x) d x)

\int_{0}^{\frac{3}{4}} 1 d x = \frac{3}{4}

\int_{0}^{\frac{3}{4}} cos (2 π x) d x = - \frac{1}{2 π}

= \frac{1}{2} (\frac{3}{4} - (- \frac{1}{2 π}))

Simplify

\frac{1}{2} (\frac{3}{4} - (- \frac{1}{2 π})) : \frac{3 π + 2}{8 π}

= \frac{3 π + 2}{8 π}

\frac{\partial}{\partial y} (2 x y - e^{2 y})

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{9})

f (x) = 3 e^{4 x}

y^{^{''}} + 25 y = 25 sec^{2} (5 x)

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + y^{\frac{1}{3}})

What is the integral from 0 to 3/4 of sin^2(pix) ?
The integral from 0 to 3/4 of sin^2(pix) is (3pi+2)/(8pi)