What is the integral from 0 to pi/4 of sin^7(y) ?

The integral from 0 to pi/4 of sin^7(y) is -(177-128sqrt(2))/(280sqrt(2))

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integral from 0 to pi/4 of sin^7(y)

\int_{0}^{\frac{π}{4}} sin^{7} (y) d y

- \frac{177 - 128 2}{280 2}

Decimal

0.01015 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sin^{7} (y) d y

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} (1 - cos^{2} (y))^{3} sin (y) d y

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - (1 - u^{2})^{3} d u

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

Expand

- (1 - u^{2})^{3} : - 1 + 3 u^{2} - 3 u^{4} + u^{6}

= \int_{1}^{\frac{2}{2}} - 1 + 3 u^{2} - 3 u^{4} + u^{6} d u

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

= - \int_{\frac{2}{2}}^{1} - 1 + 3 u^{2} - 3 u^{4} + u^{6} 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} 3 u^{2} d u - \int_{\frac{2}{2}}^{1} 3 u^{4} d u + \int_{\frac{2}{2}}^{1} u^{6} d u)

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

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

\int_{\frac{2}{2}}^{1} 3 u^{4} d u = 3 (\frac{1}{5} - \frac{1}{20 2})

\int_{\frac{2}{2}}^{1} u^{6} d u = \frac{1}{7} - \frac{1}{56 2}

= - (- (1 - \frac{1}{2}) + 1 - \frac{1}{2 2} - 3 (\frac{1}{5} - \frac{1}{20 2}) + \frac{1}{7} - \frac{1}{56 2})

Join

- (1 - \frac{1}{2}) + 1 - \frac{1}{2 2} - 3 (\frac{1}{5} - \frac{1}{20 2}) + \frac{1}{7} - \frac{1}{56 2} : \frac{177 - 128 2}{280 2}

= - \frac{177 - 128 2}{280 2}

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What is the integral from 0 to pi/4 of sin^7(y) ?
The integral from 0 to pi/4 of sin^7(y) is -(177-128sqrt(2))/(280sqrt(2))