Question

Evaluate the following indefinite integral:

∫sin^2(4x)cos^2(4x)dx


(If possible, say which identities/formulas you are using)

Answer 1

start by using\[\sin^2(u)=\frac12(1-\cos(2u))\]with\[u=4x\]

Answer 2

\[\sin^{2}(4x)\cos^{2}(4x)=\frac{ 1 }{ 4 }(2\sin(4x)\cos(4x))^{2}=\frac{ 1 }{ 4 }\sin^{2}(8x)\]But\[\sin^{2}(8x)=\frac{ 1-\cos(16x) }{ 2 }\]Thus\[\int\limits \sin^{2}(4x)\cos^{2}.dx = \int\limits \frac{ 1 }{ 8 }(1-\cos(16x)).dx\]
\[=\frac{ 1 }{ 8 }(x-\frac{ 1 }{ 16 }\sin(16x))\]

Answer 3

@zekarias . im sorry. why is: \[1/4(2\sin(4x)\cos(4x))^2=1/4\sin^2(8x)\]
this is mainly the part i don't understand.

Answer 4

2sin(x)cos(x)=sin(2x)

Answer 5

oh forgot that identity...thanks I see it now!