What is the integral from 0 to pi/2 of 15sin^7(x) ?

The integral from 0 to pi/2 of 15sin^7(x) is 48/7

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integral from 0 to pi/2 of 15sin^7(x)

\int_{0}^{\frac{π}{2}} 15 sin^{7} (x) d x

\frac{48}{7}

Decimal Notation

6.85714 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 15 sin^{7} (x) d x

Take the constant out:

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

= 15 \cdot \int_{0}^{\frac{π}{2}} sin^{7} (x) d x

Rewrite using trig identities

= 15 \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

= 15 \cdot \int_{1}^{0} - (1 - u^{2})^{3} d u

Expand

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

= 15 \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

= 15 (- \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

= 15 (- (- \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}

= 15 (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7}))

Simplify

15 (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7})) : \frac{48}{7}

= \frac{48}{7}

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What is the integral from 0 to pi/2 of 15sin^7(x) ?
The integral from 0 to pi/2 of 15sin^7(x) is 48/7