Question:

<p>Consider $I = \int \cos \sqrt{x} \, dx$</p><p>Let $\sqrt{x} = t$ then $\frac{1}{2\sqrt{x}} \, dx = dt$</p><p>i.e. $dx = 2\sqrt{x} \, dt$ or $dx = 2t \, dt$.</p><p>So $I = \int \cos t \cdot 2t \, dt$</p><p>Taking $t$ as first function, integrating it by parts, we get</p><p>$I = 2 \left[ t \int \cos t \, dt - \int \left\{ \frac{dt}{dt} \int \cos t \, dt \right\} dt \right]$</p><p>$\Rightarrow I = 2 \left[ t \sin t - \int 1 \cdot \sin t \, dt \right] = 2 [t \sin t + \cos t] + C$</p><p>$\Rightarrow I = 2 [\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$</p>