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

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

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

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

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

Solution steps

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

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} (x) d x

Rewrite using trig identities

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

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 x) d x

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 x + \int cos (2 x) d x)

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{3 π})

sin^{3} (2 t)

f (x) = - 3 x^{2} + 4 x - 1

x \to 2 lim (\frac{( x ^{2} - 5 x + 4 )}{x - 2})

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{2}} d x

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