calculus - Integrate $int sin(sqrt{at}),dt$

Integrate $$\int \sin(\sqrt{at})\,dt$$

Here is what I tried.

Let $u=\sqrt{at}$, then $\displaystyle\ du=\frac{a}{2\sqrt{at}}dt=\frac{a}{2u}dt\implies \frac{2udu}{a}=dt.$ So by subsitution,

$$\displaystyle \int \sin(\sqrt{at})dt=\int\sin(u)\left(\frac{2udu}{a}\right)=\frac{2}{a}\int u \sin(u)du.$$

Again substituting, $v=u\implies dv=du, dw=\sin(u)du\implies w=-\cos(u)$. So

$$\displaystyle\begin{align} \frac{2}{a}\int\ u\sin(u)du &= -u\cos(u)+\int\cos(u)du\\
&=-u\cos(u)+\sin(u)+C\\
&=-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})+C\\

\int u\sin(u)du&=\frac{a}{2}\left(-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})\right).\\
\end{align}$$

But the answer is $\displaystyle\frac{2}{a}\left(-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})\right)$. Where did I go wrong here?

Answer

Your mistake is when you write
$$\displaystyle\begin{align} \frac{2}{a}\int\ u\sin(u)du &= -u\cos(u)+\int\cos(u)du\\
&=-u\cos(u)+\sin(u)+C\\
&=-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})+C\\
\int u\sin(u)du&=\frac{a}{2}\left(-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})\right).\\

\end{align}$$ It is rather
$$\displaystyle\begin{align} \frac{2}{a}\int\ u\sin(u)du &= \color{red}{\frac{2}{a}}\left(-u\cos(u)+\int\cos(u)du\right)\\
&=\color{red}{\frac{2}{a}}\left(-u\cos(u)+\sin(u)+C\right)\\
&=\color{red}{\frac{2}{a}}\left(-\sqrt{at}\cos(\sqrt{at})+\sin(\sqrt{at})+C\right)
\end{align}
$$ giving at the end the right answer.