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

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

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integral of-4sin(2t-pi/2)

\int - 4 sin (2 t - \frac{π}{2}) d t

2 sin (2 t) + C

Solution steps

\int - 4 sin (2 t - \frac{π}{2}) d t

Take the constant out:

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

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

Rewrite using trig identities

= - 4 \cdot \int - cos (2 t) d t

Take the constant out:

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

= - 4 (- \int cos (2 t) d t)

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

= - 4 (- \int cos (u) \frac{1}{2} d u)

Take the constant out:

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

= - 4 (- \frac{1}{2} \cdot \int cos (u) d u)

Use the common integral:

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

= - 4 (- \frac{1}{2} sin (u))

Substitute back

u = 2 t

= - 4 (- \frac{1}{2} sin (2 t))

Simplify

- 4 (- \frac{1}{2} sin (2 t)) : 2 sin (2 t)

= 2 sin (2 t)

Add a constant to the solution

= 2 sin (2 t) + C

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

f (x) = \frac{1}{2 x + 3}

y^{^{'}} + \frac{2}{x} y = \frac{1}{x ^{2}}

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

y^{^{'}} = 2 (t + 1) y, y (0) = 1

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