## University Calculus: Early Transcendentals (3rd Edition)

$$\int\cos\sqrt xdx=2(\sqrt x\sin \sqrt x+\cos\sqrt x)+C$$
$$A=\int\cos\sqrt xdx$$ We set $a=\sqrt x$, which means $$da=\frac{1}{2\sqrt x}dx=\frac{1}{2a}dx$$ $$dx=2ada$$ Therefore, $$A=2\int a\cos ada$$ Next, we do integration by parts. Take $u=a$ and $dv=\cos ada$ We would have $du=da$ and $v=\sin a$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=2(a\sin a-\int \sin ada)$$ $$A=2(a\sin a+\cos a)+C$$ $$A=2(\sqrt x\sin \sqrt x+\cos\sqrt x)+C$$