hi i am running into trouble using trig subsitution in this particular problem so far the farthest that i have got to is
1/8 times the intergral of 5/2 - 1/2cos(2x)+ 3/4cos(4x)+ 1/2 cos(4x)cos(2x)
i really appreacated the help and thank you
hi i am running into trouble using trig subsitution in this particular problem so far the farthest that i have got to is
1/8 times the intergral of 5/2 - 1/2cos(2x)+ 3/4cos(4x)+ 1/2 cos(4x)cos(2x)
i really appreacated the help and thank you
$\displaystyle \int\sin^{4}(x)\cos^{2}(x)dx$
$\displaystyle \int\left(\sin(x)\cos(x)\right)^{2}\sin^{2}(x)dx$
$\displaystyle \int\left(\frac{\sin(2x)}{2}\right)^{2}\left(\frac {1-\cos(2x)}{2}\right)dx$
$\displaystyle \frac{1}{8}\int\sin^{2}(2x)\left(1-\cos(2x)\right)dx$
$\displaystyle \frac{1}{8}\int\left(\sin^{2}(2x)-\sin^{2}(2x)\cos(2x)\right)dx$
$\displaystyle \frac{1}{8}\int\left[\left(\frac{1-\cos(4x)}{2}\right)-\sin^{2}(2x)\cos(2x)\right]dx$
$\displaystyle \frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(4x)}{2}-\sin^{2}(2x)\cos(2x)\right]dx$
$\displaystyle \frac{1}{8}\left[\frac{x}{2}-\frac{\sin(4x)}{8}-\frac{1}{6}\sin^{3}(2x)\right]$