What is the integral of 16sin^2(x)cos^2(x) ?

The integral of 16sin^2(x)cos^2(x) is 2(x-1/4 sin(4x))+C

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integral of 16sin^2(x)cos^2(x)

\int 16 sin^{2} (x) cos^{2} (x) d x

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

Solution steps

\int 16 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

= 16 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

16 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : 2 (x - \frac{1}{4} sin (4 x))

= 2 (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= 2 (x - \frac{1}{4} sin (4 x)) + C

\int_{0}^{5} (7 x - x^{2}) - (2 x) d x

x \to \infty lim (5 + \frac{6}{x})

n = 0 \sum \infty 4^{- n}

\frac{d}{d y} (y cos (x) + 2 x e^{y})

2 y^{^{''}} - 4 y^{^{'}} + 8 y = 0

What is the integral of 16sin^2(x)cos^2(x) ?
The integral of 16sin^2(x)cos^2(x) is 2(x-1/4 sin(4x))+C