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

The integral of (cos^2(t))/2 is 1/4 (t+1/2 sin(2t))+C

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integral of (cos^2(t))/2

integral

\frac{cos ^{2} ( t )}{2}

\frac{1}{4} (t + \frac{1}{2} sin (2 t)) + C

Solution steps

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

Rewrite using trig identities

= \frac{1}{2} \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

= \frac{1}{2} \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

= \frac{1}{2} \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)

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{x}{( x + 1 )}

3 - 2 x

\frac{2}{( 3 + e ^{x} )}

\frac{sin ( y )}{cos ^{2} ( y )}

What is the integral of (cos^2(t))/2 ?
The integral of (cos^2(t))/2 is 1/4 (t+1/2 sin(2t))+C