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

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

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

integral

\frac{sin ^{2} ( x )}{2}

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

Solution steps

\int \frac{sin ^{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 sin^{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

in t e g r a l \frac{1}{( e ^{x} + e ^{- x} )}

in t e g r a l \frac{x}{( x + 2 )}

in t e g r a l (e^{- t})^{2}

in t e g r a l 2 x (x^{2} + 1)^{\frac{1}{2}}

in t e g r a l \frac{1}{- x}

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