Thursday, December 30, 2021

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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