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Integrate the following functions: sin^4x |
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Answer» Solution :`sin^4x = (SINX)^2 = ((1-cos2x)/2)^2` = `1/4(1-2cos 2x +cos^2 2x)` =`1/4(1-2cos 2x +(1+cos 4x)/2)` `=1/8(2-4 cos2x+1+cos4x)` `=1/8(3-4 cos2x+cos4x)` therefore` INT sin^4x dx` `=1/8[3x-4 (SIN2X)/2+(sin4x)/4]+c` `=3/8x- 1/4sin2x + (sin4x)/(32)+c` |
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