What is the integral of 8sin^2(x)cos^2(x) ?

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

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integral of 8sin^2(x)cos^2(x)

\int 8 sin^{2} (x) cos^{2} (x) d x

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

Solution steps

\int 8 sin^{2} (x) cos^{2} (x) d x

Take the constant out:

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

= 8 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 8 \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

= 8 \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

= 8 \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)

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{1}{x ^{6}}, (2, \frac{1}{64})

\int 1 x^{3} d x

f (x) = \frac{6 x}{x ^{2} + 1}, (0, 0)

\int \frac{1}{48} x d x

f (x) = (x - 1)^{2} + 1, at x = 0

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