What is the integral from 0 to pi/2 of 1 (sin^7(x)) ?

The integral from 0 to pi/2 of 1 (sin^7(x)) is 16/35 d

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integral from 0 to pi/2 of 1 (sin^7(x))

\int_{0}^{\frac{π}{2}} \frac{d}{1} (sin^{7} (x)) d x

\frac{16}{35} d

Solution steps

\int_{0}^{\frac{π}{2}} \frac{d}{1} sin^{7} (x) d x

Take the constant out:

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

= d \cdot \int_{0}^{\frac{π}{2}} sin^{7} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - (1 - u^{2})^{3} d u

= d \cdot \int_{1}^{0} - (1 - u^{2})^{3} d u

Expand

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

= d \cdot \int_{1}^{0} - 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

= d (- \int_{0}^{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

= d (- (- \int_{0}^{1} 1 d u + \int_{0}^{1} 3 u^{2} d u - \int_{0}^{1} 3 u^{4} d u + \int_{0}^{1} u^{6} d u))

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} 3 u^{2} d u = 1

\int_{0}^{1} 3 u^{4} d u = \frac{3}{5}

\int_{0}^{1} u^{6} d u = \frac{1}{7}

= d (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7}))

Simplify

d (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7})) : \frac{16}{35} d

= \frac{16}{35} d

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What is the integral from 0 to pi/2 of 1 (sin^7(x)) ?
The integral from 0 to pi/2 of 1 (sin^7(x)) is 16/35 d