What is the integral of 1-cos^2(x) ?

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

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integral of 1-cos^2(x)

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

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

Solution steps

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

Rewrite using trig identities

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

Add a constant to the solution

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

\int \frac{1}{x ^{1.5}} d x

f (x) = 2 cos (sin (x))

\frac{d y}{d t} - \frac{7}{t} y = t^{5}

f (x) = (x^{2} - 5 x + 8) (3 x^{2} - 2)

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

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