What is the integral of 1-cos(2t) ?

The integral of 1-cos(2t) is t-1/2 sin(2t)+C

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integral of 1-cos(2t)

\int 1 - cos (2 t) d t

t - \frac{1}{2} sin (2 t) + C

Solution steps

\int 1 - cos (2 t) d t

Apply the Sum Rule:

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

= \int 1 d t - \int cos (2 t) d t

\int 1 d t = t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

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

Add a constant to the solution

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

y = x^{3} - 2 x - 2, at x = 1

- sin (x)

\frac{d}{d x} (x^{p} e^{a x})

\int \frac{y}{1 - y ^{2}} d y

\frac{\partial}{\partial x} (- 1 + ln (x))

What is the integral of 1-cos(2t) ?
The integral of 1-cos(2t) is t-1/2 sin(2t)+C