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

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

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integral from 0 to pi of sin^4(3x)

\int_{0}^{π} sin^{4} (3 x) d x

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

\int_{0}^{π} sin^{4} (3 x) d x

Apply u-substitution

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

= \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

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

Apply Integral Reduction

= \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}

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

Simplify

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

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

Compute the boundaries:

0

= \frac{1}{3} (- 0 + \frac{9 π}{8})

Simplify

= \frac{3 π}{8}

\int - 1 d x

\frac{\partial}{\partial x} (cos (2 x - 3 y))

\int cos (x) \cdot sin (2 x) d x

\frac{\partial}{\partial x} (cos (2 x y))

3 x^{2} y d x + x^{3} d y = 0

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