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

The integral from 0 to pi/4 of sin(4t) is 1/2

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integral from 0 to pi/4 of sin(4t)

\int_{0}^{\frac{π}{4}} sin (4 t) d t

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{\frac{π}{4}} sin (4 t) d t

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{4} \cdot 2

Simplify

= \frac{1}{2}

d er i v a t i v e ln (\frac{x - 2}{x + 2})

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{8})

f (x) = sin^{3} (3 x)

f (t) = cos (2 t)

\frac{d}{d x} (2 \cdot cos (x))

What is the integral from 0 to pi/4 of sin(4t) ?
The integral from 0 to pi/4 of sin(4t) is 1/2