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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{\frac{π}{4}} sin (4 x) d x

Apply u-substitution

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

= \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}

x^{3} - 12 x^{2} + 20 x, - x^{3} + 12 x^{2} - 20 x

f (x) = 7 tan (x)

\int \frac{1}{1 + csc ( x )} d x

\int e^{x} \cdot cos (2 x) d x

7 x

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