Integrate.....

Compute the definite integral:
 integral_0^π sin^2(x) cos^2(x) dx
Write cos^2(x) as 1 - sin^2(x):
 = integral_0^π sin^2(x) (1 - sin^2(x)) dx
Expanding the integrand sin^2(x) (1 - sin^2(x)) gives sin^2(x) - sin^4(x):
 = integral_0^π (sin^2(x) - sin^4(x)) dx
Integrate the sum term by term and factor out constants:
 = - integral_0^π sin^4(x) dx + integral_0^π sin^2(x) dx
Use the reduction formula, integral sin^m(x) dx = -(cos(x) sin^(m - 1)(x))/m + (m - 1)/m integral sin^(-2 + m)(x) dx, where m = 4:
 = 1/4 sin^3(x) cos(x) right bracketing bar _0^π + 1/4 integral_0^π sin^2(x) dx
Evaluate the antiderivative at the limits and subtract.
 1/4 sin^3(x) cos(x) right bracketing bar _0^π = (1/4 sin^3(π) cos(π)) - 1/4 sin^3(0) cos(0) = 0:
 = 1/4 integral_0^π sin^2(x) dx
Write sin^2(x) as 1/2 - 1/2 cos(2 x):
 = 1/4 integral_0^π (1/2 - 1/2 cos(2 x)) dx
Integrate the sum term by term and factor out constants:
 = -1/8 integral_0^π cos(2 x) dx + 1/8 integral_0^π 1 dx
For the integrand cos(2 x), substitute u = 2 x and du = 2 dx.
This gives a new lower bound u = 2 0 = 0 and upper bound u = 2 π:
 = -1/16 integral_0^(2 π) cos(u) du + 1/8 integral_0^π 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of cos(u) is sin(u):
 = (-(sin(u))/16) right bracketing bar _0^(2 π) + 1/8 integral_0^π 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-(sin(u))/16) right bracketing bar _0^(2 π) = (-1/16 sin(2 π)) - (-(sin(0))/16) = 0:
 = 1/8 integral_0^π 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
 = x/8 right bracketing bar _0^π
Evaluate the antiderivative at the limits and subtract.
 x/8 right bracketing bar _0^π = π/8 - 0/8 = π/8:
Answer: | = π/8

Like so:

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Integrate the following with steps please:

integrate [sin^2 (x) cos^2 (x)] dx from x=0 to pi.

Thank you for any help.

Formula:

\(\begin{array}{|rcll|} \hline \cos^2(x) = \frac12\cdot \Big(1+\cos(2x)\Big) \\ \sin^2(x) = \frac12\cdot \Big(1-\cos(2x)\Big) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \int \limits_{0}^{\pi} {\sin^2(x) \cos^2(x)\ dx} \\ &=& \int \limits_{0}^{\pi} {\frac12\cdot \Big(1-\cos(2x)\Big)\cdot \frac12\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos(2x)\Big)\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos^2(2x)\Big) \ dx} \quad & | \quad \cos^2(2x) = \frac12\cdot \Big(1+\cos(4x)\Big) \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12\cdot \Big(1+\cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big( \frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot [\frac{x}{2}-\frac{\sin(4x)}{4}]_{0}^{\pi} \\ &=& \frac18\cdot [ x - \frac{\sin(4x)}{2} ]_{0}^{\pi} \\ &=& \frac18\cdot [ \pi - \sin(4\pi)-(0-\frac{\sin(4\cdot 0)}{2} ) ] \\ &=& \frac18\cdot [ \pi - 0-(0-0) ] \\ &=& \frac18\cdot [ \pi ] \\ &=& \frac{\pi}{8}\\ \hline \end{array}\)

*edited