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Integration of sec^4(2x)dx |
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Answer» USE the trigonometric identities: #sin^2 ALPHA = (1-cos2alpha)/2# #cos^2 alpha = (1+cos2alpha)/2# to have: #sin^4(2X) = (sin^2(2x))^2 = (1-cos(4x))^2/4# #sin^4(2x) = (1-2cos(4x) +cos^2(4x))/4# #sin^4(2x) = 1/4-(cos(4x))/2 +1/8 +(cos(8x))/8# #sin^4(2x) = 3/8-(cos(4x))/2 +(cos(8x))/8# Using the linearity of the integral: #int sin^4(2x) dx= 3/8int dx-1/2 int cos(4x)dx +1/8 INTCOS(8x)dx# #int sin^4(2x) dx= (3x)/8-1/8 int cos(4x)d(4x) +1/64 intcos(8x)d(8x)# #int sin^4(2x) dx= (3x)/8-(sin(4x))/8 + (sin(8x))/64+C# |
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