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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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