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

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

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integral of-2sin^2(x)

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

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

Solution steps

\int - 2 sin^{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 sin^{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

\frac{d y}{d x} + 2 x y = 4 x

\int \frac{10}{( 25 x ^{2} + 1 ) ^{2}} d x

\int_{1}^{5} 7 r^{2} ln (r) d r

x \to - 2 lim (\frac{3}{x ^{2} - 4})

\int \frac{x - 2}{x ^{3}} d x

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