Question

Please help me finish the following integration problem (click to see).

Answer 1

\[\int\limits_{}^{}\sin^2xcos^2xdx\]

Answer 2

this is where i am stuck...
\[\frac{ x }{ 4 }-\frac{ 1 }{ 4 }\int\limits_{}^{}\cos^2(2x)dx\]

Answer 3

okay use u sub

Answer 4

You can always use the power reduction identity. cos^2(x)=1/2(1+cos(2x))

Answer 5

i mean by parts

Answer 6

@Babyslapmafro how did u get their

Answer 7

\[\int\limits_{}^{}\frac{ 1-\cos^2(2x) }{ 4 }\]

Answer 8

u=1/4
v=\[\int\limits_{}^{}\cos^2(2x)dx\]

Answer 9

are those the correct values for u and v?

Answer 10

hmm use \[\cos^2x= -\sin^2x+1\]

Answer 11

did u get the idea

Answer 12

\[\int\limits_{}^{}\sin^2x(-\sin^2x+1)\]

Answer 13

\[\int\limits_{}^{}-\sin^4x+\sin^2x\]

Answer 14

lol what i have done