1.

Integrate `sinx.sin2x.sin3x+sec^2x*cos^2 2x+sin^4x*cos^4x`

Answer» Correct Answer - `("cos"4x)/(16)-("cos"2x)/(8)+("cos"6x)/(24)+"sin"2x + "tan" x - 2x + (3x)/(128)-("sin"4x)/(128)+("sin"8x)/(1024)`
Let `I_(1)=int sin x sin 2x sin 3"x dx"`
`(1)/(4)int(sin 4x+sin 2x - sin 6x)dx`
`=-(cos 4x)/(16)-(cos 2x)/(8)+(cos 6x)/(24)`
`I_(2)=int sec^(2)x*cos^(2)2xdx`
`=int sec^(2)x(2 cos^(2)x-1)^(2)dx`
`=int(4 cos^(2)x+sec^(2)x-4)dx`
`=int(2 cos 2x + sec^(2)x x-2)dx`
`=sin 2x + tan x -2x and I_(3)=int sin^(4)x cos^(4)xdx`
`=(1)/(128)int(3-4 cos 4x + cos 8x)dx`
`=(3x)/(128)-(sin 4x)/(128)+(sin 8x)/(1024)`
`therefore" "I=I_(1)+I_(2)+I_(3)`
`=-(cos 4x)/(16)-(cos 2x)/(8)+(cos 6x)/(24)+sin 2x + tan x-2x + (3x)/(128)-(sin 4x)/(128)+(sin 8x)/(1024)`


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