What is the integral from 0 to pi/5 of sin^4(5x) ?

The integral from 0 to pi/5 of sin^4(5x) is (3pi)/(40)

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integral from 0 to pi/5 of sin^4(5x)

\int_{0}^{\frac{π}{5}} sin^{4} (5 x) d x

\frac{3 π}{40}

Decimal

0.23561 \dots

Solution steps

\int_{0}^{\frac{π}{5}} sin^{4} (5 x) d x

Apply u-substitution

: \int_{0}^{π} sin^{4} (u) \frac{1}{5} d u

= \int_{0}^{π} sin^{4} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{π} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{5} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (u) d u)

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

= \frac{1}{5} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

\frac{1}{5} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : \frac{1}{5} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

= \frac{1}{5} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= \frac{1}{5} (- 0 + \frac{3 π}{8})

Simplify

= \frac{3 π}{40}

y^{^{'}} + y = \frac{1}{1 + t ^{2}}

x \to 0 lim (x^{2} ln (4 x^{2}))

\frac{1}{4 s ^{3}}

\frac{\partial}{\partial x} (1 - x^{2} - x - 3 y^{2} + y)

\frac{\partial}{\partial x} (x y - 4 x^{2} y)

What is the integral from 0 to pi/5 of sin^4(5x) ?
The integral from 0 to pi/5 of sin^4(5x) is (3pi)/(40)