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

The integral from 0 to pi/2 of 1sin^2(x) is pi/4

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integral from 0 to pi/2 of 1sin^2(x)

\int_{0}^{\frac{π}{2}} 1 sin^{2} (x) d x

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 1 \cdot sin^{2} (x) d x

Rewrite using trig identities

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

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

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

= \frac{1}{2} (\frac{π}{2} - 0)

Simplify

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

= \frac{π}{4}

\int_{- 4}^{0} x^{3} - x^{2} - 14 x d x

\int_{0}^{2} x + 5 d x

\int_{π}^{x} t^{3} sin^{2} (t) d t

\int_{0}^{2} \frac{y ^{3}}{1 + y ^{4}} d y

\int_{2}^{3} (2 x + 5) d x

What is the integral from 0 to pi/2 of 1sin^2(x) ?
The integral from 0 to pi/2 of 1sin^2(x) is pi/4