12 Days of Christmas Integrals: ∫ cos (√x) dx

12 Days of Christmas Integrals:  ∫ cos (√x) dx

On the Sixth day of Christmas Integrals, the integral featured today is...

∫ cos (√x) dx

Handling this integral will require two integral methods.  First substitution:

Let u = √x = x^(1/2)

Then:

du = 1/2 ∙ x^(-1/2) dx

2 ∙ x^(1/2) du = dx

2 ∙ u du = dx

∫ cos(√x) dx

= ∫ cos(x^(1/2)) dx

= ∫ 2 ∙ u ∙ cos(u) du

At this point, we now apply Integration by Parts:

w = 2 ∙ u

dw = 2 du

dv = cos(u) du

v = sin(u)

= 2 ∙ u ∙ sin(u) - ∫ 2 ∙ sin(u) du

= 2 ∙ u ∙ sin(u) +  2 ∙ cos(u) + C

Recall u = x^(1/2):

= 2 ∙ x^(1/2) ∙ sin(x^(1/2)) + 2 ∙ cos(x^(1/2)) + C

Eddie 

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