Integration of (Sin2xco2x/√4-sin^4)dx
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Answer:
38x−18sin(4x)+164sin(8x)+C
Explanation:
∫sin4(2x)dx
=∫(sin2(2x))2dx
=∫(1−cos(4x)2)2dx
=14∫(1−2cos(4x)+cos2(4x))dx
=14∫(1−2cos(4x)+1+cos(8x)2)dx
=18∫(3−4cos(4x)+cos(8x))dx
=18(3x−sin(4x)+18sin(8x))+C
=38x−18sin(4x)+164sin(8x)+C
38x−18sin(4x)+164sin(8x)+C
Explanation:
∫sin4(2x)dx
=∫(sin2(2x))2dx
=∫(1−cos(4x)2)2dx
=14∫(1−2cos(4x)+cos2(4x))dx
=14∫(1−2cos(4x)+1+cos(8x)2)dx
=18∫(3−4cos(4x)+cos(8x))dx
=18(3x−sin(4x)+18sin(8x))+C
=38x−18sin(4x)+164sin(8x)+C
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