How do you evaluate the integral #int sinsqrtx/sqrtx#?

2 Answers
Noah G
Mar 5, 2017

#-2cossqrt(x) + C#

Explanation:

Let #u = sqrt(x)#. Then #du = 1/(2sqrt(x)) dx# and #dx = 2sqrtxdu#.

#int sinu/u * 2udu#

#2int sinu du#

#-2cosu + C#

#-2cossqrt(x) + C#

Hopefully this helps!

mason m
Jun 22, 2017

#-2cossqrtx+C#

Explanation:

The following is an absolutely ridiculous way of arriving at the correct answer. I only did this method because this question was filed under "Trigonometric Substitutions", so I used a trig function in my substitution.

While this worked, using #u=sqrtx# is much more straightforward.

Let #x=sin^2theta#. Then #dx=2sinthetacosthetad theta# and #sqrtx=sintheta#.

#I=intsinsqrtx/sqrtxdx=intsin(sin theta)/sintheta2sinthetacosthetad theta#

#color(white)I=2intsin(sin theta)costhetad theta#

Letting #t=sintheta# shows that #dt=costhetad theta#:

#I=2intsintdt=-2cost=-2cos(sin theta)=-2cossqrtx+C#

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