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

Mar 5, 2017

$- 2 \cos \sqrt{x} + C$

#### Explanation:

Let $u = \sqrt{x}$. Then $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ and $\mathrm{dx} = 2 \sqrt{x} \mathrm{du}$.

$\int \sin \frac{u}{u} \cdot 2 u \mathrm{du}$

$2 \int \sin u \mathrm{du}$

$- 2 \cos u + C$

$- 2 \cos \sqrt{x} + C$

Hopefully this helps!

Jun 22, 2017

$- 2 \cos \sqrt{x} + C$

#### 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 $u = \sqrt{x}$ is much more straightforward.

Let $x = {\sin}^{2} \theta$. Then $\mathrm{dx} = 2 \sin \theta \cos \theta d \theta$ and $\sqrt{x} = \sin \theta$.

$I = \int \sin \frac{\sqrt{x}}{\sqrt{x}} \mathrm{dx} = \int \sin \frac{\sin \theta}{\sin} \theta 2 \sin \theta \cos \theta d \theta$

$\textcolor{w h i t e}{I} = 2 \int \sin \left(\sin \theta\right) \cos \theta d \theta$

Letting $t = \sin \theta$ shows that $\mathrm{dt} = \cos \theta d \theta$:

$I = 2 \int \sin t \mathrm{dt} = - 2 \cos t = - 2 \cos \left(\sin \theta\right) = - 2 \cos \sqrt{x} + C$