What is the integral of sin^2(9x)cos^2(9x) ?

The integral of sin^2(9x)cos^2(9x) is 1/8 (x-1/36 sin(36x))+C

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integral of sin^2(9x)cos^2(9x)

\int sin^{2} (9 x) cos^{2} (9 x) d x

\frac{1}{8} (x - \frac{1}{36} sin (36 x)) + C

Solution steps

\int sin^{2} (9 x) cos^{2} (9 x) d x

Rewrite using trig identities

= \int \frac{1}{8} (1 - cos (36 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 (36 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 (36 x) d x)

\int 1 d x = x

\int cos (36 x) d x = \frac{1}{36} sin (36 x)

= \frac{1}{8} (x - \frac{1}{36} sin (36 x))

Add a constant to the solution

= \frac{1}{8} (x - \frac{1}{36} sin (36 x)) + C

What is the integral of sin^2(9x)cos^2(9x) ?
The integral of sin^2(9x)cos^2(9x) is 1/8 (x-1/36 sin(36x))+C