integration of cos^2[3x]
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I= ∫cos²(3x) dx
We know cos2x = 2cos²x-1
=> Cos²x= (cos2x+1)/2
Here cos²(3x) = (cos6x+1)/2
= 1/2∫(cos6x+1 )dx
{ since ∫m{f(x)}dx = m∫f(x) dx where m is constant
= 1/2∫cos6x dx + ∫ dx
= 1/2{(sin6x/6) +(x)} + c
{ since ∫cosax dx = 1/a sinax+c and ∫ 1.dx = x +c
We know cos2x = 2cos²x-1
=> Cos²x= (cos2x+1)/2
Here cos²(3x) = (cos6x+1)/2
= 1/2∫(cos6x+1 )dx
{ since ∫m{f(x)}dx = m∫f(x) dx where m is constant
= 1/2∫cos6x dx + ∫ dx
= 1/2{(sin6x/6) +(x)} + c
{ since ∫cosax dx = 1/a sinax+c and ∫ 1.dx = x +c
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