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

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

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

\int_{0}^{\frac{π}{2}} sin (4 x) d x

0

Solution steps

\int_{0}^{\frac{π}{2}} sin (4 x) d x

Apply u-substitution

: \int_{0}^{2 π} sin (u) \frac{1}{4} d u

= \int_{0}^{2 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}^{2 π} sin (u) d u

Use the common integral:

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

= \frac{1}{4} [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{4} \cdot 0

Simplify

= 0

\int_{0}^{\frac{π}{2}} f (sin (x)) d x

\int_{0}^{1} \frac{x}{1 - x} d x

\int_{- 1}^{1} (1 + 2 x + 3 x^{2}) d x

\int_{e}^{e^{2}} \frac{ln ( x )}{x} d x

\int_{1}^{x} sin (t^{2}) d t

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