\(\int\limits8sin^4(x)dx =\)
Take the constant out \(\int\limits a*f(x)dx=a* \int\limitsf(x)dx\)
\(= 8 \int\limits sin^4(x)dx\)
\(\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx\)
\(= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)\)
Use the following identify : \(sin^2(x) = \frac{1-cos(2x)}{2}\)
\(= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)\)
\(\int\limits 1dx = x\)
\(\int\limits cos(2x)dx = \frac{1}{2} sin(2x)\)
Apply the sum rule :
\(=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))\)
Simplify :
\(8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))\)
Therefore add a constant to the solution:
\(= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C\)
hope this helps!
