What is the integral of-3cos^2(x)sin^2(x) ?

The integral of-3cos^2(x)sin^2(x) is -3/8 (x-1/4 sin(4x))+C

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integral of-3cos^2(x)sin^2(x)

\int - 3 cos^{2} (x) sin^{2} (x) d x

- \frac{3}{8} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int - 3 cos^{2} (x) sin^{2} (x) d x

Take the constant out:

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

= - 3 \cdot \int cos^{2} (x) sin^{2} (x) d x

Rewrite using trig identities

= - 3 \cdot \int \frac{1 - cos ( 4 x )}{8} d x

Take the constant out:

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

= - 3 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= - 3 \cdot \frac{1}{8} (\int 1 d x - \int cos (4 x) d x)

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= - 3 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

- 3 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : - \frac{3}{8} (x - \frac{1}{4} sin (4 x))

= - \frac{3}{8} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= - \frac{3}{8} (x - \frac{1}{4} sin (4 x)) + C

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\frac{\partial}{\partial z} (e^{x y} + z)

\frac{d}{d x} ((x + 2)^{2} + (10 - 3 x)^{2})

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What is the integral of-3cos^2(x)sin^2(x) ?
The integral of-3cos^2(x)sin^2(x) is -3/8 (x-1/4 sin(4x))+C