What is #int cos^3(x) sin^4(x) dx #?

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Answer 1 · 2018-04-27T01:07:52

The integral is equal to #(7sin^5x-5sin^7x)/35+C#.

Use #cos^2x=1-sin^2x#, then use a substitution:

#color(white)=intcos^3x*sin^4x# #dx#

#=intcosx*cos^2x*sin^4x# #dx#

#=intcosx*(1-sin^2x)*sin^4x# #dx#

Let #u=sinx#, so #du=cosx# #dx#, and #dx=(du)/cosx#:

#=intcosx*(1-u^2)*u^4\*(du)/cosx#

#=intcolor(red)cancelcolor(black)cosx*(1-u^2)*u^4\*(du)/color(red)cancelcolor(black)cosx#

#=int(1-u^2)*u^4# #du#

#=int(u^4-u^6)# #du#

#=intu^4# #du-intu^6# #du#

#=u^5/5-u^7/7+C#

#=(7u^5-5u^7)/35+C#

#=(7sin^5x-5sin^7x)/35+C#

That's the integral. Hope this helped!