What is the integral of 3cos^2(t) ?

The integral of 3cos^2(t) is 3/2 (t+1/2 sin(2t))+C

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integral of 3cos^2(t)

\int 3 cos^{2} (t) d t

\frac{3}{2} (t + \frac{1}{2} sin (2 t)) + C

Solution steps

\int 3 cos^{2} (t) d t

Take the constant out:

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

= 3 \cdot \int cos^{2} (t) d t

Rewrite using trig identities

= 3 \cdot \int \frac{1 + cos ( 2 t )}{2} d t

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \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

= 3 \cdot \frac{1}{2} (\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)

= 3 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t))

Simplify

3 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t)) : \frac{3}{2} (t + \frac{1}{2} sin (2 t))

= \frac{3}{2} (t + \frac{1}{2} sin (2 t))

Add a constant to the solution

= \frac{3}{2} (t + \frac{1}{2} sin (2 t)) + C

\frac{d ^{2} y}{d x ^{2}} + 9 y = 0

6 x^{3} - tan (x)

\frac{\partial}{\partial x} (x y (1 - 7 x - 4 y))

8 cos (2 x), 8 - 8 cos (2 x), 0, \frac{π}{2}

x \to 0 + lim (x - \frac{1}{x ^{3}})

What is the integral of 3cos^2(t) ?
The integral of 3cos^2(t) is 3/2 (t+1/2 sin(2t))+C