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#