# Thread: Integral 07

1. ## Integral 07

$\displaystyle \int \frac{sec(\sqrt{x})dx}{\sqrt{x}}$

2. Don't you have any thought on this one?

3. Integration by parties?

$\displaystyle u=\sqrt{x}$
$\displaystyle v=\int sec(\sqrt{x})dx$

4. No, u substitution. Let $\displaystyle u = \sqrt x$ ...

5. but in denominator is $\displaystyle sec(\sqrt{x})$

6. Just do what I said and you'll see why I did it..

Let $\displaystyle u = \sqrt x$... Now what's du ?

7. $\displaystyle du=\frac{2x^\frac{3}{2}}{3}$

8. $\displaystyle \int \frac{sec(\sqrt{x})dx}{\sqrt{x}}=\int {2\sec (u)du}\;\;,\; u=\sqrt{x}$

9. Why?
$\displaystyle =\int 2sec(u)du$

10. Surely you see it.

If we let $\displaystyle u=\sqrt{x}$, then $\displaystyle du=\frac{1}{2\sqrt{x}}dx$

and $\displaystyle 2du=\frac{1}{\sqrt{x}}dx$

See?. The 2du replaces the $\displaystyle \frac{1}{\sqrt{x}}dx$ in your integral.

Then, we get $\displaystyle 2\int sec(u)du$

11. Originally Posted by Apprentice123
Why?
$\displaystyle =\int 2sec(u)du$
You make the substitution $\displaystyle u=\sqrt{x}\implies\,du=\frac{\,dx}{2\sqrt{x}}$

$\displaystyle \therefore\int\frac{\sec(\sqrt{x})}{\sqrt{x}}\,dx\ implies 2\int\frac{\sec(u)}{2\sqrt{x}}\,dx\implies 2\int\sec(u)\,du$

--Chris

12. What kind of integral you used?