What is the integral from 0 to pi/6 of (cos(3x))^8 ?

The integral from 0 to pi/6 of (cos(3x))^8 is (35pi)/(768)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/6 of (cos(3x))^8

\int_{0}^{\frac{π}{6}} (cos (3 x))^{8} d x

\frac{35 π}{768}

Decimal Notation

0.14317 \dots

Solution steps

\int_{0}^{\frac{π}{6}} (cos (3 x))^{8} d x

Simplify

= \int_{0}^{\frac{π}{6}} cos^{8} (3 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos^{8} (u) \frac{1}{3} d u

= \int_{0}^{\frac{π}{2}} cos^{8} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} cos^{8} (u) d u

Apply Integral Reduction

= \frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \int_{0}^{\frac{π}{2}} cos^{6} (u) d u)

\int_{0}^{\frac{π}{2}} cos^{6} (u) d u = \frac{5 π}{32}

= \frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32})

Simplify

\frac{1}{3} ([\frac{sin ( u ) cos ^{7} ( u )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32}) : \frac{1}{3} ([\frac{1}{8} cos^{7} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{35 π}{256})

= \frac{1}{3} ([\frac{1}{8} cos^{7} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{35 π}{256})

Compute the boundaries:

0

= \frac{1}{3} (0 + \frac{35 π}{256})

Simplify

= \frac{35 π}{768}

\int_{0}^{\frac{π}{6}} sin^{4} (x) d x

\int_{0}^{1} (2^{x} - 2) d x

\int_{- 1}^{0} 384 x^{5} (8 x^{6} - 3)^{3} d x

\int_{0}^{1} (2 x + 1)^{5} d x

\int_{0}^{3} (5 + 3 cos (x)) d x

What is the integral from 0 to pi/6 of (cos(3x))^8 ?
The integral from 0 to pi/6 of (cos(3x))^8 is (35pi)/(768)