How do you find the antiderivative of #int (sin(sqrtx)) dx#?

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Answer 1 · 2017-01-26T21:55:41

#intsin(sqrtx)dx=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C#

#I=intsin(sqrtx)dx#

Let #t=sqrtx#. This implies that #x=t^2# so #dx=2tdt#.

Then:

#I=intsin(t)(2tdt)=int2tsin(t)dt#

Now we will use integration by parts. Let:

#{(u=2t,=>,du=2dt),(dv=sin(t)dt,=>,v=-cos(t)):}#

Then:

#I=uv-intvdu#

#I=-2tcos(t)-int(-cos(t))2dt#

#I=-2tcos(t)+2intcos(t)#

#I=-2tcos(t)+2sin(t)+C#

From #t=sqrtx#:

#I=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C#