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

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

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integral of (cos(t))^2

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

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

Solution steps

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

Simplify

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Add a constant to the solution

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

x \to 0 lim (\frac{sin ( 3 x )}{sin ( 6 x )})

\int 3 e^{x} - 9 cos (x) + C d x

4 x^{3}

\frac{d y}{d x} = tan (x) + sec (x)

\frac{cos ( x )}{2 sin ( x )}

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