What is the integral of $int sin^2 (x).cos^2 (x) dx$ ?

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Answer 1 · 2016-06-19T06:02:41

$intsin^2xcos^2xdx=x/8-(sin4x)/32+c$

As $sin2x=2sinxcosx$

$intsin^2xcos^2xdx=1/4int(4sin^2xcos^2x)dx$

= $1/4intsin^2(2x)dx$

= $1/4int(1-cos4x)/2dx$

= $x/8-1/8intcos4xdx$

= $x/8-1/8xx(sin4x)/4+c$

= $x/8-(sin4x)/32+c$

What is the integral of int sin^2 (x).cos^2 (x) dx int sin^2 (x).cos^2 (x) dx ?