Question

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

Answer 1

`-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!

Answer 2

`-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`