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

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

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integral of-2cos^2(x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

= - 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

= - 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

= - 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)

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 2 sec (3 x^{2})

(6 y + 2 t - 7) d t + (8 y + 6 t - 1) d y = 0, y (- 1) = 2

\int_{0}^{t} t^{2} e^{- \frac{t}{2}} d t

\int \frac{1}{x ^{4} - x} d x

\frac{d}{d x} (\frac{2 x ^{2} tan ( x )}{sec ( x )})

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