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

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

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

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

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

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

Rewrite using trig identities

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\frac{d}{d x} (- 2 x^{2} - 5)

y = x^{2} - 2, y = 2

2 x^{2} + y^{2} = 9

5 t^{2} ln (t)

e^{x}, e^{- 4 x}, ln (3)

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