What is the integral from 0 to pi/2 of cos^3(31x) ?

The integral from 0 to pi/2 of cos^3(31x) is -2/93

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integral from 0 to pi/2 of cos^3(31x)

\int_{0}^{\frac{π}{2}} cos^{3} (31 x) d x

- \frac{2}{93}

Decimal Notation

- 0.02150 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{3} (31 x) d x

Apply u-substitution

: \int_{0}^{\frac{31 π}{2}} cos^{3} (u) \frac{1}{31} d u

= \int_{0}^{\frac{31 π}{2}} cos^{3} (u) \frac{1}{31} d u

Take the constant out:

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

= \frac{1}{31} \cdot \int_{0}^{\frac{31 π}{2}} cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{31} \cdot \int_{0}^{\frac{31 π}{2}} (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int_{0}^{- 1} 1 - v^{2} d v

= \frac{1}{31} \cdot \int_{0}^{- 1} 1 - v^{2} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{31} (- \int_{- 1}^{0} 1 - v^{2} 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}{31} (- (\int_{- 1}^{0} 1 d v - \int_{- 1}^{0} v^{2} d v))

\int_{- 1}^{0} 1 d v = 1

\int_{- 1}^{0} v^{2} d v = \frac{1}{3}

= \frac{1}{31} (- (1 - \frac{1}{3}))

Simplify

\frac{1}{31} (- (1 - \frac{1}{3})) : - \frac{2}{93}

= - \frac{2}{93}

x \to 5 - lim (5)

x \to - 2 lim (x^{2} + 9 + 14)

f (x) = ln (6 - x^{2} + 2 x^{4}), at x = 1

\frac{d y}{d x} + (\frac{3}{x}) y = - x + 4

\frac{\partial}{\partial x} (ln (\frac{1}{2 + 3 e ^{x}}))

What is the integral from 0 to pi/2 of cos^3(31x) ?
The integral from 0 to pi/2 of cos^3(31x) is -2/93