Question #108784

∫cos²(ax)dx


1
Expert's answer
2020-04-14T13:47:47-0400

Answer: cos2(ax)  dx=12x+sin  2ax4a+C\int\cos^2(ax)\;dx=\frac12x+\frac{\sin\;2ax}{4a}+C

cos2(ax)  dx=  1aa  cos2(ax)  dxu=ax    du  =  a  dxSubstitute  u.1aa  cos2(ax)  dx=1acos2u  ducos2u=  1+cos  2u2cos2u  du  =  (12+cos  2u2)  du  =  =12u+12×sin2u2+C=12u+sin2u4+C==12ax+sin  2ax4+Ccos2(ax)  dx  =1a(  12ax+sin  2ax4)+C==12x+sin  2ax4a+C\int\cos^2(ax)\;dx=\;\frac1a\int a\;\cos^2(ax)\;dx\\u=ax\;\Rightarrow\;du\;=\;a\;dx\\Substitute\;u.\\\frac1a\int a\;\cos^2(ax)\;dx=\frac1a\int\cos^2u\;du\\\cos^2u=\;\frac{1+\cos\;2u}2\\\int\cos^2u\;du\;=\;\int(\frac12+\frac{\cos\;2u}2)\;du\;=\;\\=\frac12u+\frac12\times\frac{\sin2u}2+C=\frac12u+\frac{\sin2u}4+C=\\=\frac12ax+\frac{\sin\;2ax}4+C\\\int\cos^2(ax)\;dx\;=\frac1a(\;\frac12ax+\frac{\sin\;2ax}4)+C=\\=\frac12x+\frac{\sin\;2ax}{4a}+C


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