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

The integral from 0 to pi/2 of sin(3x) is 1/3

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integral from 0 to pi/2 of sin(3x)

\int_{0}^{\frac{π}{2}} sin (3 x) d x

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin (3 x) d x

Apply u-substitution

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

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

Use the common integral:

\int sin (u) d u = - cos (u)

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

Compute the boundaries:

1

= \frac{1}{3} \cdot 1

Simplify

= \frac{1}{3}

\frac{d}{d x} (3^{\frac{1}{2}})

n \to \infty lim (e^{- n} ln (n))

\int x + 3 d x

y^{^{'}} = sin (x + y) + cos (x + y)

\int (x^{2} + 2 x) e^{x} d x

What is the integral from 0 to pi/2 of sin(3x) ?
The integral from 0 to pi/2 of sin(3x) is 1/3