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

The integral from 0 to pi/4 of sin^2(x) is (pi-2)/8

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integral from 0 to pi/4 of sin^2(x)

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

\frac{π - 2}{8}

Decimal Notation

0.14269 \dots

Solution steps

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

Rewrite using trig identities

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

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

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

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

Simplify

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

= \frac{π - 2}{8}

f (t) = \frac{1}{t}

\int (csc^{2} (x)) d x

f (x) = 4 x - x^{2}

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

\int x^{20} e^{- x^{21}} d x

What is the integral from 0 to pi/4 of sin^2(x) ?
The integral from 0 to pi/4 of sin^2(x) is (pi-2)/8