What is the integral of 16cos^2(x+3)sin^2(x+3) ?

The integral of 16cos^2(x+3)sin^2(x+3) is 2(x-1/4 sin(12+4x))+C

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integral of 16cos^2(x+3)sin^2(x+3)

\int 16 cos^{2} (x + 3) sin^{2} (x + 3) d x

2 (x - \frac{1}{4} sin (12 + 4 x)) + C

Solution steps

\int 16 cos^{2} (x + 3) sin^{2} (x + 3) d x

Take the constant out:

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

= 16 \cdot \int cos^{2} (x + 3) sin^{2} (x + 3) d x

Rewrite using trig identities

= 16 \cdot \int \frac{1 - cos ( 4 ( 3 + 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 (3 + x)) d x

Expand

4 (3 + x) : 12 + 4 x

= 16 \cdot \frac{1}{8} \cdot \int 1 - cos (12 + 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 (12 + 4 x) d x)

\int 1 d x = x

\int cos (12 + 4 x) d x = \frac{1}{4} sin (12 + 4 x)

= 16 \cdot \frac{1}{8} (x - \frac{1}{4} sin (12 + 4 x))

Simplify

16 \cdot \frac{1}{8} (x - \frac{1}{4} sin (12 + 4 x)) : 2 (x - \frac{1}{4} sin (12 + 4 x))

= 2 (x - \frac{1}{4} sin (12 + 4 x))

Add a constant to the solution

= 2 (x - \frac{1}{4} sin (12 + 4 x)) + C

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What is the integral of 16cos^2(x+3)sin^2(x+3) ?
The integral of 16cos^2(x+3)sin^2(x+3) is 2(x-1/4 sin(12+4x))+C