$\int \frac{1 - cos 3 x}{1 - cos x}$ $int (1-cos 3x)/(1-cos x)$ ?

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Answer 1 · 2018-05-31T05:25:39

$sin 2 x + 4 sin x + 3 x + C$ $sin2x+4sinx+3x+C$ .

Since, $cos 3 x - 1 = (4 {cos}^{3} x - 3 cos x) - 1$ $cos3x-1=(4cos^3x-3cosx)-1$ , we find that,

$cos 3 x - 1$ $cos3x-1$ is a cubic polynomial in $cos x$ $cosx$ , and noting that the sum

of its co-efficients being $0, (cos x - 1)$ $0, (cosx-1)$ is a factor.

Thus, $cos 3 x - 1 = 4 {cos}^{3} x - 3 cos x - 1$ $cos3x-1=4cos^3x-3cosx-1$ ,

$=ul(4cos^3x-4cos^2x)+ul(4cos^2x-4cosx)+ul(cosx-1)$ ,

$=4cos^2x(cosx-1)+4cosx(cosx-1)+1(cosx-1)$ ,

$=(cosx-1)(4cos^2x+4cosx+1)$

$=(cosx-1){2(1+cos2x)+4cosx+1}$ ,

$:. (cos3x-1)=(cosx-1)(2cos2x+4cosx+3)$ .

$rArr (cos3x-1)/(cosx-1)=(2cos2x+4cosx+3)$ .

$:. int(1-cos3x)/(1-cosx)dx=int(cos3x-1)/(cosx-1)dx$ ,

$=int(2cos2x+4cosx+3)dx$ ,

$=2*(sin2x)/2+4sinx+3x$ .

$rArr int(1-cos3x)/(1-cosx)dx=sin2x+4sinx+3x+C$ .

Feel the Joy of Maths.!

int (1-cos 3x)/(1-cos x)∫1−cos3x1−cosxint (1-cos 3x)/(1-cos x) ?