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Question

Evaluate:sin4xcos4xdx


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Solution

To evaluate : sin4xcos4xdx

Let I=sin4x.cos4x.dx

We know that

sin2x=(1-cos2x)2 and cos2x=(1+cos2x)2

Now sin4xcos4xdx=sin4xcos4xdx

=[(1-cos2x2)2(1+cos2x2)2]dx

=(1-cos2x4)2(1+cos2x4)2]dx

=116(1-cos22x)2dx

=116(1+cos42x-2cos22x)dx

Solving for cos2x , we get

cos4x=2cos22x-12cos22x=1+cos4xcos22x=1+cos4x2

sin4xcos4xdx =116(1+(1+cos24x2)2-2(1+cos4x2)dx

=116(1+(1+cos24x2)2-1-cos4xdx

=1161+cos24x+2cos4x-2cos4x2dx

=1321+cos24xdx

=1321+cos24xdx

Solving for cos4x , we get

cos8x=2cos24x-12cos24x=1+cos8xcos24x=1+cos8x2

sin4xcos4xdx =1321+1+cos8x2dx

=1322+1+cos8x2dx

=1643+cos8xdx

=164[3+cos8x]dx

=164[3x+sin8x8]+C (Formula used : =cosnx=sinnxn)

Thus,sin4xcos4xdx=164[3x+sin8x8]+C


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