What is the integral of+sin^2(6x)cos^2(6x) ?

The integral of+sin^2(6x)cos^2(6x) is 1/8 (x-1/24 sin(24x))+C

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integral of+sin^2(6x)cos^2(6x)

\int + sin^{2} (6 x) cos^{2} (6 x) d x

\frac{1}{8} (x - \frac{1}{24} sin (24 x)) + C

Solution steps

\int sin^{2} (6 x) cos^{2} (6 x) d x

Rewrite using trig identities

= \int \frac{1}{8} (1 - cos (24 x)) 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 (24 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 (24 x) d x)

\int 1 d x = x

\int cos (24 x) d x = \frac{1}{24} sin (24 x)

= \frac{1}{8} (x - \frac{1}{24} sin (24 x))

Add a constant to the solution

= \frac{1}{8} (x - \frac{1}{24} sin (24 x)) + C

\int [\frac{1}{x ^{2}} + sec^{2} (π x)] d x

\frac{\partial}{\partial x} ((x y)^{\frac{1}{2}})

n = 8 \sum \infty \frac{1}{4 ^{n}}

\int \frac{x ^{3}}{( x + 1 ) ^{3}} d x

\frac{d}{d x} (2 x (x + 4)^{3})

What is the integral of+sin^2(6x)cos^2(6x) ?
The integral of+sin^2(6x)cos^2(6x) is 1/8 (x-1/24 sin(24x))+C