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

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

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

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

\frac{π - 2}{8}

Solution steps

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

Simplify

= \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}

\frac{d}{d x} (5 lo g_{10} (4 x^{2}))

\int csc^{3} (4 x) d x

y^{^{'}} = y (x y^{2} + 3)

\frac{d}{d x} (\frac{ln ( x ^{3} )}{ln ( x ^{2} )})

(3, - 8), (- 6, - 3)

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