1. ## intergral of sin^4(x)cos^2(x)

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

2. Originally Posted by momo55
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

How did you get to your result? Please try to show some work. Also, you will have to use the reduction formula for integration. Read this. It might help.

that you'll benefit from sin^2(x) = 1 - cos^2(x) along with integration by parts.

4. $\int\sin^{4}(x)\cos^{2}(x)dx$

$\int\left(\sin(x)\cos(x)\right)^{2}\sin^{2}(x)dx$

$\int\left(\frac{\sin(2x)}{2}\right)^{2}\left(\frac {1-\cos(2x)}{2}\right)dx$

$\frac{1}{8}\int\sin^{2}(2x)\left(1-\cos(2x)\right)dx$

$\frac{1}{8}\int\left(\sin^{2}(2x)-\sin^{2}(2x)\cos(2x)\right)dx$

$\frac{1}{8}\int\left[\left(\frac{1-\cos(4x)}{2}\right)-\sin^{2}(2x)\cos(2x)\right]dx$

$\frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(4x)}{2}-\sin^{2}(2x)\cos(2x)\right]dx$

$\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(4x)}{8}-\frac{1}{6}\sin^{3}(2x)\right]$