What is the integral from 0 to pi of 3sin^4(3t) ?

The integral from 0 to pi of 3sin^4(3t) is (9pi)/8

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 of 3sin^4(3t)

\int_{0}^{π} 3 sin^{4} (3 t) d t

\frac{9 π}{8}

Decimal Notation

3.53429 \dots

Solution steps

\int_{0}^{π} 3 sin^{4} (3 t) d t

Take the constant out:

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

= 3 \cdot \int_{0}^{π} sin^{4} (3 t) d t

Apply u-substitution

: \int_{0}^{3 π} sin^{4} (u) \frac{1}{3} d u

= 3 \cdot \int_{0}^{3 π} sin^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{3 π} sin^{4} (u) d u

Apply Integral Reduction

= 3 \cdot \frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{3 π} + \frac{3}{4} \cdot \int_{0}^{3 π} sin^{2} (u) d u)

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

= 3 \cdot \frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{3 π} + \frac{3}{4} \cdot \frac{3 π}{2})

Simplify

3 \cdot \frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{3 π} + \frac{3}{4} \cdot \frac{3 π}{2}) : - [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{3 π} + \frac{9 π}{8}

= - [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{3 π} + \frac{9 π}{8}

Compute the boundaries:

0

= - 0 + \frac{9 π}{8}

Simplify

= \frac{9 π}{8}

\frac{15}{( s + 10 ) ( s + 11 )}

\int_{0}^{2.8} 0.2 x d x

\frac{4 π r ^{3}}{3}

\frac{x ^{3}}{x + 1}

\int \frac{1}{x ^{p}} d x

What is the integral from 0 to pi of 3sin^4(3t) ?
The integral from 0 to pi of 3sin^4(3t) is (9pi)/8