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

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

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integral of sin^2(x/b)cos^2(x/b)

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

\frac{1}{8} (x - \frac{b}{4} sin (\frac{4 x}{b})) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{8} (1 - cos (\frac{4 x}{b})) 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 (\frac{4 x}{b}) 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 (\frac{4 x}{b}) d x)

\int 1 d x = x

\int cos (\frac{4 x}{b}) d x = \frac{b}{4} sin (\frac{4 x}{b})

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

Add a constant to the solution

= \frac{1}{8} (x - \frac{b}{4} sin (\frac{4 x}{b})) + C

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