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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Add a constant to the solution

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

(2 y^{2} + 3 x y) + (2 x y + x^{2}) y^{^{'}} = 0

x^{^{''}} + 9 x = 0, x (0) = 0.7, x^{^{'}} (0) = 0.4

x \to 2 lim (- x^{2})

\int 7 x 1 - x^{4} d x

\frac{d}{d x} (\frac{1}{2 x - 5})

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