What is the integral of-16sin(4x-pi/2) ?

The integral of-16sin(4x-pi/2) is 4sin(4x)+C

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integral of-16sin(4x-pi/2)

\int - 16 sin (4 x - \frac{π}{2}) d x

4 sin (4 x) + C

Solution steps

\int - 16 sin (4 x - \frac{π}{2}) d x

Take the constant out:

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

= - 16 \cdot \int sin (4 x - \frac{π}{2}) d x

Rewrite using trig identities

= - 16 \cdot \int - cos (4 x) d x

Take the constant out:

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

= - 16 (- \int cos (4 x) d x)

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= - 16 (- \int cos (u) \frac{1}{4} d u)

Take the constant out:

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

= - 16 (- \frac{1}{4} \cdot \int cos (u) d u)

Use the common integral:

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

= - 16 (- \frac{1}{4} sin (u))

Substitute back

u = 4 x

= - 16 (- \frac{1}{4} sin (4 x))

Simplify

- 16 (- \frac{1}{4} sin (4 x)) : 4 sin (4 x)

= 4 sin (4 x)

Add a constant to the solution

= 4 sin (4 x) + C

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

\int \frac{21 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

y^{^{'}} = 3 y, y (2) = 4

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

\int x^{2} + k x + 1 d x

What is the integral of-16sin(4x-pi/2) ?
The integral of-16sin(4x-pi/2) is 4sin(4x)+C