Integrate
Cos^7(2x)
Explanation would be helpful
2 answers
Let u=2x du/dx=2 =1/2Incos^7u(int=integral) 1/2Intcos^6u*cosu since cos^2u=1-sin^2u there cos^6u=(cos^2u)^3=(1-sin^2u)^3 there the integral become:1/2Int(1-sin^2u)^3cosudu let v=sinu dv/du=cosu du=dv/cosu substitute it in du and v in integral gives:1/2Int(1-v^2)^3dv expansion of (1-v^2)^3=1-3v^2+3v^4-v^6 so the integral becomes:In 1-In3v^2+In3v^4-Inv^6 integrating each gives:v-v^3+3/5v^5-v^7/7 substitute v=sinu:sinu-sin^3u+3/5sin^5u-1/7sin^7u substitute u=2x:sin2x-sin^3(2x)+3/5sin^5(2x)-1/7sin^7(2x)
wow! what a lot of work.
juat factor out the cosx and expand the (1-sin^2 x)^3
juat factor out the cosx and expand the (1-sin^2 x)^3
1 answer