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

The integral from-pi/4 to pi/4 of cos^7(2x) is 16/35

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integral from-pi/4 to pi/4 of cos^7(2x)

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{7} (2 x) d x

\frac{16}{35}

Decimal

0.45714 \dots

Solution steps

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{7} (2 x) d x

Apply u-substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{7} (u) \frac{1}{2} d u

= \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{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_{- \frac{π}{2}}^{\frac{π}{2}} cos^{7} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

\int_{- 1}^{1} 1 d v = 2

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

\int_{- 1}^{1} 3 v^{4} d v = \frac{6}{5}

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

= \frac{1}{2} (2 - 2 + \frac{6}{5} - \frac{2}{7})

Simplify

\frac{1}{2} (2 - 2 + \frac{6}{5} - \frac{2}{7}) : \frac{16}{35}

= \frac{16}{35}

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What is the integral from-pi/4 to pi/4 of cos^7(2x) ?
The integral from-pi/4 to pi/4 of cos^7(2x) is 16/35