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

x \to 0 lim (x^{3 x})

\int_{1}^{2} \frac{4 + u ^{2}}{u ^{3}} d u

\frac{d y}{d x} = \frac{x ^{2}}{y}

\frac{d x}{d t} = 2 (x^{2} + 1), x (\frac{π}{4}) = 1

\int \frac{x ^{2} - 2 x - 1}{x} d x

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