What is the integral from 0 to pi/4 of sin^7(2x) ?

The integral from 0 to pi/4 of sin^7(2x) is 8/35

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integral from 0 to pi/4 of sin^7(2x)

\int_{0}^{\frac{π}{4}} sin^{7} (2 x) d x

\frac{8}{35}

Decimal

0.22857 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sin^{7} (2 x) d x

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} sin^{7} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= \frac{1}{2} \cdot \int_{1}^{0} - 1 + 3 v^{2} - 3 v^{4} + v^{6} d v

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

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

Apply the Sum Rule:

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

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

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

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

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

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

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

Simplify

\frac{1}{2} (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7})) : \frac{8}{35}

= \frac{8}{35}

What is the integral from 0 to pi/4 of sin^7(2x) ?
The integral from 0 to pi/4 of sin^7(2x) is 8/35