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Question

Find sin4xdx.


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Solution

Find the integration of the given function.

sin4xdx=(sin2x)2dx

=1cos2x22dx...[sin2x=1-cos2x2]=14(1cos2x)2dx=14(12cos2x+cos22x)dx...[(a-b)2=a2-2ab+b2]=1412cos2x+1+cos4x2dx=14dx24cos2xdx+18(1+cos4x)dx=x412sin2x2+x8+18sin4x4+C

sin4xdx=38x+132sin(4x)14sin(2x)+C

Hence, the required answer is sin4xdx=38x+132sin(4x)14sin(2x)+C


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