What is the integral of 1-cos^2(θ) ?

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

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integral of 1-cos^2(θ)

\int 1 - cos^{2} (θ) d θ

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

Solution steps

\int 1 - cos^{2} (θ) d θ

Rewrite using trig identities

= \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 \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} (\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} (θ - \frac{1}{2} sin (2 θ))

Add a constant to the solution

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

\frac{d}{d x} (7 x + 2)

y^{^{'}} + y = e^{- t} ln (t)

\frac{d}{d x} (\frac{ln ( x + y )}{ln ( y )})

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

\frac{4}{( s ( s ^{2} + 4 ) )}

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