What is the integral from 0 to 1 of sin^3(pi/2 x) ?

The integral from 0 to 1 of sin^3(pi/2 x) is 4/(3pi)

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integral from 0 to 1 of sin^3(pi/2 x)

\int_{0}^{1} sin^{3} (\frac{π}{2} x) d x

\frac{4}{3 π}

Decimal Notation

0.42441 \dots

Solution steps

\int_{0}^{1} sin^{3} (\frac{π}{2} x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{c o s (1)} - \frac{sin ^{3} ( \frac{π a r c c o s ( u )}{2} )}{1 - u ^{2}} d u

= \int_{1}^{c o s (1)} - \frac{sin ^{3} ( \frac{π a r c c o s ( u )}{2} )}{1 - u ^{2}} d u

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

= - \int_{c o s (1)}^{1} - \frac{sin ^{3} ( \frac{π a r c c o s ( u )}{2} )}{1 - u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - - \int_{c o s (1)}^{1} \frac{sin ^{3} ( \frac{π a r c c o s ( u )}{2} )}{1 - u ^{2}} d u

Apply u-substitution

: \int_{1}^{0} - sin^{3} (\frac{π v}{2}) d v

= - (- \int_{1}^{0} - sin^{3} (\frac{π v}{2}) d v)

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

= - (- (- \int_{0}^{1} - sin^{3} (\frac{π v}{2}) d v))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- (- (- \int_{0}^{1} sin^{3} (\frac{π v}{2}) d v)))

Apply Integration By Parts

: [sin^{3} (\frac{π v}{2}) v - \int \frac{3 π}{2} sin^{2} (\frac{π v}{2}) cos (\frac{π v}{2}) v d v]_{0}^{1}

= - (- (- (- [sin^{3} (\frac{π v}{2}) v - \int \frac{3 π}{2} sin^{2} (\frac{π v}{2}) cos (\frac{π v}{2}) v d v]_{0}^{1})))

\int \frac{3 π}{2} sin^{2} (\frac{π v}{2}) cos (\frac{π v}{2}) v d v = v sin^{3} (\frac{π v}{2}) - \frac{2}{3 π} (cos^{3} (\frac{π v}{2}) - 3 cos (\frac{π v}{2}))

= - (- (- (- [sin^{3} (\frac{π v}{2}) v - (v sin^{3} (\frac{π v}{2}) - \frac{2}{3 π} (cos^{3} (\frac{π v}{2}) - 3 cos (\frac{π v}{2})))]_{0}^{1})))

Simplify

- (- (- (- [sin^{3} (\frac{π v}{2}) v - (v sin^{3} (\frac{π v}{2}) - \frac{2}{3 π} (cos^{3} (\frac{π v}{2}) - 3 cos (\frac{π v}{2})))]_{0}^{1}))) : [\frac{2}{3 π} cos^{3} (\frac{π v}{2}) - \frac{2}{π} cos (\frac{π v}{2})]_{0}^{1}

= [\frac{2}{3 π} cos^{3} (\frac{π v}{2}) - \frac{2}{π} cos (\frac{π v}{2})]_{0}^{1}

Compute the boundaries:

\frac{4}{3 π}

= \frac{4}{3 π}

x \to 0 lim (3 x^{2} + 1)

\int_{0}^{x} t (e^{2 t^{3}} - 1) d t

\frac{d y}{d x}, x^{3} - x y + y^{2} = 7

\frac{d}{d x} (\frac{x - 1}{x})

x \to 0 lim (\frac{9}{x} - \frac{9}{x ^{2} - x})

What is the integral from 0 to 1 of sin^3(pi/2 x) ?
The integral from 0 to 1 of sin^3(pi/2 x) is 4/(3pi)