+0  
 
0
1
1066
2
avatar

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+26402 
+30

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

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 

\(\begin{array}{lcl} \sin^3{(x)} \cos^5{(x)} && \\ = \sin{(x)}\cdot \sin^2{(x)}\cos^5{(x)} \qquad | \qquad \sin^2{(x)} = 1-\cos^2{(x)}\\ = \sin{(x)}\cdot [ {1-\cos^2{(x)}} ] \cdot \cos^5{(x)} \\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^2{(x)}\cos^5{(x)}} ] \qquad | \qquad \cos^2{(x)}\cos^5{(x)} = \cos^7{(x)}\\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^7{(x)}} ] \end{array}\)

 

laugh

 Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
edited by heureka  Oct 22, 2015
 #1
avatar+26402 
+30
Best Answer

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

they then rewrite it as:

sin x (cos^5(x) - cos^7(x))

how?

 

\(\begin{array}{lcl} \sin^3{(x)} \cos^5{(x)} && \\ = \sin{(x)}\cdot \sin^2{(x)}\cos^5{(x)} \qquad | \qquad \sin^2{(x)} = 1-\cos^2{(x)}\\ = \sin{(x)}\cdot [ {1-\cos^2{(x)}} ] \cdot \cos^5{(x)} \\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^2{(x)}\cos^5{(x)}} ] \qquad | \qquad \cos^2{(x)}\cos^5{(x)} = \cos^7{(x)}\\ = \sin{(x)}\cdot [ {\cos^5{(x)}-\cos^7{(x)}} ] \end{array}\)

 

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

3 Online Users

avatar