Science:Math Exam Resources/Courses/MATH101/April 2008/Question 03 (b)/Solution 1

As in the hint, let $u = \sqrt{x}$ so that $d u = \frac{d x}{2 \sqrt{x}} = \frac{d x}{2 u}$ . This gives

$\int \cos \sqrt{x} d x = \int 2 u \cos (u) d u$

To solve this last integral, we use integration by parts. Let

$\begin{aligned} w = u & v = \sin (u) \\ d w = d u & d v = \cos (u) d u \end{aligned}$

Then, we have

$\begin{aligned} \int 2 u \cos (u) d u & = 2 \int u \cos (u) d u \\ = 2 u \sin (u) - 2 \int \sin (u) d u \\ = 2 u \sin (u) + 2 \cos (u) + C \end{aligned}$

and hence

$\begin{aligned} \int \cos (\sqrt{x}) d x & = 2 \sqrt{x} \sin (\sqrt{x}) + 2 \cos (\sqrt{x}) + C . \end{aligned}$