text: # Evaluate the integral  integrate of (sin(x))^2(cos(x))^4 dx
  
  jeew-m | Certified Educator
  
  calendarEducator since 2012
  
  starTop subjects are Math, Science, and Social Sciences
  
  `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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explanation: The document contains a math question and a detailed answer to the question. The question is about evaluating the integral of \((\sin(x))^2(\cos(x))^4 \, dx\), and the answer provides a step-by-step solution to the integral.
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distilabel_metadata.content: # Evaluate the integral  integrate of (sin(x))^2(cos(x))^4 dx
  
  jeew-m | Certified Educator
  
  calendarEducator since 2012
  
  starTop subjects are Math, Science, and Social Sciences
  
  `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`
  
  check Approved by eNotes Editorial
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer to the question. The question is about evaluating the integral of \((\sin(x))^2(\cos(x))^4 \, dx\), and the answer provides a step-by-step solution to the integral.
  
  Label: question-and-answer