How do you integrate #int sin(sqrtx)# by integration by parts method?

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Answer 1 · 2016-09-02T17:19:26

# 2sin(sqrt(x))-2 sqrt(x) cos(sqrt(x))+C#

Making #x = y^2# in #int sin(sqrtx)dx#

after #dx = 2 y dy# we have

#int sin(sqrtx)dx equiv 2inty sin y dy#

but #d/(dy)(y cos y) = cosy -y sin y# so

#2inty sin y dy=2int cos y dy -2y cos y = 2sin y -2y cos y + C#

Finally

#int sin(sqrtx)dx = 2sin(sqrt(x))-2 sqrt(x) cos(sqrt(x))+C#