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

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

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

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

\frac{π - 2}{8}

Decimal Notation

0.14269 \dots

Solution steps

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

Rewrite using trig identities

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

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 t) d t

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 t - \int_{0}^{\frac{π}{4}} cos (2 t) d t)

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

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

\int_{0}^{2} \frac{x}{4} d x

\int_{1}^{e} \frac{1}{x ( 2 x + 1 )} d x

\int_{a}^{b} 3 - x^{2} d x

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

\int_{3}^{4} 1

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