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

The integral from 0 to pi/4 of sin^3(x) is 2/3-5/(6sqrt(2))

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integral from 0 to pi/4 of sin^3(x)

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

\frac{2}{3} - \frac{5}{6 2}

Decimal

0.07741 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

= \int_{1}^{\frac{2}{2}} - 1 + u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{2}{2}}^{1} - 1 + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int_{\frac{2}{2}}^{1} 1 d u + \int_{\frac{2}{2}}^{1} u^{2} d u)

\int_{\frac{2}{2}}^{1} 1 d u = 1 - \frac{1}{2}

\int_{\frac{2}{2}}^{1} u^{2} d u = \frac{1}{3} - \frac{1}{6 2}

= - (- (1 - \frac{1}{2}) + \frac{1}{3} - \frac{1}{6 2})

Simplify

- (- (1 - \frac{1}{2}) + \frac{1}{3} - \frac{1}{6 2}) : \frac{2}{3} - \frac{5}{6 2}

= \frac{2}{3} - \frac{5}{6 2}

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

16 y^{^{''}} + 8 y^{^{'}} + y = x + 6

y (x + 5) + y^{^{'}} = 0

n = 1 \sum \infty \frac{x ^{n}}{n ^{n}}

x \to 0 lim (- (\frac{1}{x}))

What is the integral from 0 to pi/4 of sin^3(x) ?
The integral from 0 to pi/4 of sin^3(x) is 2/3-5/(6sqrt(2))