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

integral

\frac{- cos ^{2} ( x )}{2}

- \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

Simplify

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

Simplify

(- \frac{1}{2}) \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

1. 3^{x}

\frac{3 x}{( x ^{2} + 4 x + 6 )}

\frac{12}{x ^{2}}

\frac{( 2 x )}{( 4 + 3 x ) ^{2}}

\frac{1}{( 4 + 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