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Evaluate the integral  integrate of (sin(x))^2(cos(x))^4 dx

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bllybony | Student | Honors

Posted October 14, 2012 at 2:26 AM via web

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Evaluate the integral  integrate of (sin(x))^2(cos(x))^4 dx

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted October 14, 2012 at 5:11 AM (Answer #1)

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`sin2x = 2sinx*cosx`

`(sin2x)^2 = 4(sinx)^2*(cos(x))^2`

`(sinx)^2*(cos(x))^2 = 1/4(sin2x)^2`

 

`cos2A = 2cos^2A-1 = 1-2sin^A`

`cos^2A = 1/2(cos2A+1)`

`sin^2A = 1/2(1-cos2A)`

 

int(sin(x))^2(cos(x))^4 dx

`= int (sinx)^2*(cosx)^2*(cosx)^2dx`

`= int (1/4(sin2x)^2)(1/2(cos2x+1))dx`

`= 1/8int [(sin2x)^2cos2x+(sin2x)^2]dx`

`= 1/8[int(sin2x)^2cos2xdx+int1/2(1-cos4x)dx]`

 

 

`int(sin2x)^2cos2xdx`

Let `t = sin2x`

`dt = 2cos2xdx`

dt/2 = cos2xdx

 

`int(sin2x)^2cos2xdx`

`= int t^2dt/2`

= 1/2(t^3/3)

`= t^3/6`

` = (sin2x)^3/6`

 

 

`int1/2(1-cos4x)dx`

`= 1/2[int 1dx-intcos4xdx]`

`= 1/2(x-1/4(sin4x))`

`= 1/8(4x-sin4x)`

 

`int(sin(x))^2(cos(x))^4 dx`

`= 1/8[int(sin2x)^2cos2xdx+int(sin2x)^2cos2xdx]`

`= 1/8[(sin2x)^3/6+1/8(4x-sin4x)]+C`

`= 1/24[4(sin2x)^3+3(4x-sin4x)]+C`

C is a constant since this is indefinite integral.

 

`int(sin(x))^2(cos(x))^4 dx = 1/24[4(sin2x)^3+3(4x-sin4x)]+C`

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