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

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

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integral of (cos^2(θ))/2

\int \frac{cos ^{2} ( θ )}{2} d θ

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

Solution steps

\int \frac{cos ^{2} ( θ )}{2} d θ

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} (θ) d θ

Rewrite using trig identities

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

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 θ) d θ

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 θ + \int cos (2 θ) d θ)

\int 1 d θ = θ

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 9 - x ^{2} )} d x

\int - 4 x sin (5 x) d x

y = 3 x + 13

x \to \infty lim (9 x^{2})

6 lo g_{9} (x)

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