What is the integral of 3sin^2(4x) ?

The integral of 3sin^2(4x) is 3/2 (x-1/8 sin(8x))+C

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integral of 3sin^2(4x)

\int 3 sin^{2} (4 x) d x

\frac{3}{2} (x - \frac{1}{8} sin (8 x)) + C

Solution steps

\int 3 sin^{2} (4 x) d x

Take the constant out:

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

= 3 \cdot \int sin^{2} (4 x) d x

Rewrite using trig identities

= 3 \cdot \int \frac{1}{2} (1 - cos (8 x)) d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 1 - cos (8 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}{2} (\int 1 d x - \int cos (8 x) d x)

\int 1 d x = x

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= 3 \cdot \frac{1}{2} (x - \frac{1}{8} sin (8 x))

Simplify

3 \cdot \frac{1}{2} (x - \frac{1}{8} sin (8 x)) : \frac{3}{2} (x - \frac{1}{8} sin (8 x))

= \frac{3}{2} (x - \frac{1}{8} sin (8 x))

Add a constant to the solution

= \frac{3}{2} (x - \frac{1}{8} sin (8 x)) + C

\frac{d y}{d x} = 2 (x + 1) - y

\frac{d}{d x} (- \frac{18}{x ^{3}} + \frac{3}{x ^{2}})

9 (t + 1) y^{^{'}} - 7 y = 14 t

2 y + sin (2 t) = \frac{d y}{d t}

\frac{d y}{d x} + y cos (x) = 4 cos (x)

What is the integral of 3sin^2(4x) ?
The integral of 3sin^2(4x) is 3/2 (x-1/8 sin(8x))+C