How do you evaluate $int sin^2(5x)cos^2(5x) dx$ for [7, 11]?

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Answer 1 · 2015-09-16T08:30:19

$1/2+1/80sin40$

Lets transform integrand using formulas:
$sin^2(x/2)=(1-cosx)/2, cos^2(x/2)=(1+cosx)/2$

In our case:

$sin^2(5x)cos^2(5x)=(1-cos10x)/2 (1+cos10x)/2=$

$=(1-cos^2(10x))/4=1/4-1/4cos^2(10x)=$

$=1/4-1/4*(1+cos20x)/2=1/4-1/8-1/8cos20x=$

$=1/8-1/8cos20x$

We have (using substitution $t=20x, dt=20dx, dx=dt/20$ for $cos20x$ part):

$int(1/8-1/8cos20x)dx = 1/8x-1/8*1/20sin20x+C=$
$=1/8x-1/160sin20x+C=I$

Changing the values we get:

$I=11/8-1/160sin220-7/8+1/160sin40=$
$=1/2-1/160(sin220-sin140)=$
$=1/2-1/160*2cos(360/2)sin(80/2)=$
$=1/2-1/80(-1)sin40=1/2+1/80sin40$

How do you evaluate int sin^2(5x)cos^2(5x) dxint sin^2(5x)cos^2(5x) dx for [7, 11]?