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