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Z sin^3(x) cos^5(x) dx

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 Oct 22, 2015

Best Answer 

 #1
avatar+26396 
+30

sin^3(x) cos^5(x)

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 

sin3(x)cos5(x)=sin(x)sin2(x)cos5(x)|sin2(x)=1cos2(x)=sin(x)[1cos2(x)]cos5(x)=sin(x)[cos5(x)cos2(x)cos5(x)]|cos2(x)cos5(x)=cos7(x)=sin(x)[cos5(x)cos7(x)]

 

laugh

 Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
 #1
avatar+26396 
+30
Best Answer

sin^3(x) cos^5(x)

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 

sin3(x)cos5(x)=sin(x)sin2(x)cos5(x)|sin2(x)=1cos2(x)=sin(x)[1cos2(x)]cos5(x)=sin(x)[cos5(x)cos2(x)cos5(x)]|cos2(x)cos5(x)=cos7(x)=sin(x)[cos5(x)cos7(x)]

 

laugh

heureka Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
 #2
avatar
+5

Z sin^3(x) cos^5(x) dx

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 

sin^3(x) cos^5(x), all these alternate forms are the same:

sin(x) cos^5(x)-sin(x) cos^7(x),

sin^7(x) cos(x)-2 sin^5(x) cos(x)+sin^3(x) cos(x),

1/128 (6 sin(2 x)+2 sin(4 x)-2 sin(6 x)-sin(8 x)),

-1/256 i (e^(-i x)-e^(i x))^3 (e^(-i x)+e^(i x))^5

 Oct 22, 2015

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