What is the integral of-6sin^2(x) ?

The integral of-6sin^2(x) is -3(x-1/2 sin(2x))+C

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integral of-6sin^2(x)

\int - 6 sin^{2} (x) d x

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

Solution steps

\int - 6 sin^{2} (x) d x

Take the constant out:

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

= - 6 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

= - 6 \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

= - 6 \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

= - 6 \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)

= - 6 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

- 6 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : - 3 (x - \frac{1}{2} sin (2 x))

= - 3 (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= - 3 (x - \frac{1}{2} sin (2 x)) + C

\int \frac{e ^{2 x}}{e ^{2 x} - 5} d x

\int 6 cos (3 x) sin (3 x) d x

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

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

x^{0}

What is the integral of-6sin^2(x) ?
The integral of-6sin^2(x) is -3(x-1/2 sin(2x))+C