What is the integral of cos(4t)-sin(4t) ?

The integral of cos(4t)-sin(4t) is 1/4 (sin(4t)+cos(4t))+C

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integral of cos(4t)-sin(4t)

\int cos (4 t) - sin (4 t) d t

\frac{1}{4} (sin (4 t) + cos (4 t)) + C

Solution steps

\int cos (4 t) - sin (4 t) d t

Apply u-substitution

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

u = 4 t

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

Simplify

= \frac{1}{4} (sin (4 t) + cos (4 t))

Add a constant to the solution

= \frac{1}{4} (sin (4 t) + cos (4 t)) + C

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

\frac{d}{d x} (e^{9 x^{2}})

x^{^{''}} + 2 x = 0

y = x^{3} - 3 x + 1

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

What is the integral of cos(4t)-sin(4t) ?
The integral of cos(4t)-sin(4t) is 1/4 (sin(4t)+cos(4t))+C