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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