What is the integral of-4sin^2(x) ?

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

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integral of-4sin^2(x)

\int - 4 sin^{2} (x) d x

- 2 x + sin (2 x) + C

Solution steps

\int - 4 sin^{2} (x) d x

Take the constant out:

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

= - 4 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

= - 4 \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

= - 4 \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

= - 4 \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)

= - 4 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

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

= - 2 x + sin (2 x)

Add a constant to the solution

= - 2 x + sin (2 x) + C

y = \frac{1}{x ^{3}}, (- 2, - \frac{1}{8})

x \to 3 + lim (\frac{1 - 2 x}{x - 3})

\frac{d}{d x} (\frac{14}{ln ( x )})

\frac{d}{d x} (cot (\frac{x}{2}))

\frac{\partial}{\partial x} (8 y - (x + y)^{3})