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

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

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

integral

(sin^{2} (x)) (cos^{2} (x))

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

Solution steps

\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

in t e g r a l (1 + x)^{4}

in t e g r a l cos^{7} (x)

in t e g r a l \frac{1}{( 1 + x ^{4} )}

in t e g r a l \frac{x ^{3}}{( x + 2 )}

in t e g r a l \frac{1}{( a ^{2} - x ^{2} )}

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