Question

Intsin^2xcos^2xdx

Answer 1

rewriting an identity might help with it

Answer 2

I tried to use the half formular but got caught in circles

Answer 3

circles are nice, there tends to be a nice way out most of the time

Answer 4

is this:\[\int sin^2(x)~cos^2(x)~dx\]

Answer 5

yes

Answer 6

well do you know the formula \[2sinx.cosx=\sin2x?\]

Answer 7

would the half formular be more appropriate here? I used 1/2(1+cos2x)

Answer 8

i wonder .... prolly taking a hard way with this but
u = cos v = 1/3 sin^3
du = -sin dv = sin^2cos dx

\[\frac13sin^3(x)cos(x)+\frac13\int sin^4(x)~dx\]
\[\frac13sin^3(x)cos(x)+\frac13\int sin^2(x)(1-cos^2(x))~dx\]
\[\frac13sin^3(x)cos(x)+\frac13\int sin^2(x)dx-\frac13\int sin^2(x)cos^2(x))~dx\]
\[\frac43\int sin^2cos^2dx=\frac13sin^3(x)cos(x)+\frac13\int sin^2(x)dx\]
\[\int sin^2cos^2dx=\frac14sin^3(x)cos(x)+\frac14\int sin^2(x)dx\]

Answer 9

yeah that is the harder way lol

Answer 10

do we know how to run sin^2 ?

Answer 11

I needed to use the process for trig functions

Answer 12

run sin^2?

Answer 13

\[cos(2x)=cos^2(x)-sin^2(x)\]
\[cos(2x)=1-sin^2(x)-sin^2(x)\]
\[cos(2x)=1-2sin^2(x)\]
\[cos(2x)-1=-2sin^2(x)\]
\[\frac12-\frac12cos(2x)=sin^2(x)\]

Answer 14

it was sin times cos not sin minus cos

Answer 15

\[\int\frac12-\frac12cos(2x)~dx\]
should run that last part

Answer 16

\[\int sin^2(x)cos^2(x)~dx=\frac14sin^3(x)cos(x)+\frac14\int \frac12-\frac12cos(2x)~dx\]
with any luck :)

Answer 17

the wolf likes it

http://www.wolframalpha.com/input/?i=d%2Fdx+%28sin%5E3x*cosx%29%2F4%2Bx%2F8-%28sin%282x%29%29%2F16

Answer 18

yeah but your doing it by parts. Theres a simpler way

Answer 19

im sure there is a simpler way; i just never really use it lol

Answer 20

\[I=\int\limits \sin ^{2}x \cos ^{2}x dx=\int\limits \frac{ 1-\cos 2x }{2 }\frac{ 1+\cos 2x }{ 2 }dx\]
\[=\frac{ 1 }{4 }\int\limits \left( 1-\cos ^{2}2x \right) dx=\frac{ 1 }{4 }\int\limits \sin ^{2}2x dx\]
\[=\frac{ 1 }{8 }\int\limits \left( 1-\cos 4x \right)dx=\frac{ 1 }{8 }\left( x-\frac{ \sin 4x }{ 4 } \right)+c\]

Answer 21

That's the way I was looking for. ty @surjithayer. and TY for your help also @amistre64