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

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

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integral of sin^2(θ)cos^2(θ)

\int sin^{2} (θ) cos^{2} (θ) d θ

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

Solution steps

\int sin^{2} (θ) cos^{2} (θ) d θ

Rewrite using trig identities

= \int \frac{1 - cos ( 4 θ )}{8} d θ

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 θ) d θ

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 θ - \int cos (4 θ) d θ)

\int 1 d θ = θ

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{1}{x} - \frac{1}{e ^{x} - 1})

\int x e d x

\frac{d}{d x} ((x + 2) e^{2 x})

\frac{\partial}{\partial x} (x ln (y) + y^{2} z + z^{2})

f (x) = 6 x - 3

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