Answer
Verify $\int\cos^2{x}dx=\frac{1}{2}x+\frac{1}{4}\sin{2x}+C$
We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral.
\begin{equation*}
\frac{d}{dx}\left(\frac{1}{2}x+\frac{1}{4}\sin{2x}+C\right)=\frac{d}{dx}\left(\frac{1}{2}x\right)+\frac{d}{dx}\left(\frac{1}{4}\sin{2x}\right)+\frac{d}{dx}\left(C\right)
\end{equation*}
\begin{equation*}
=\left(\frac{1}{2}\right)+\left(\frac{1}{4}\cos{2x}\times2\right)+\left(0\right)
\end{equation*}
\begin{equation*}
=\frac{1}{2}+\frac{1}{2}\cos{2x}=\frac{1+\cos{2x}}{2}=\cos^2{x}
\end{equation*}
as desired.
Work Step by Step
Verify $\int\cos^2{x}dx=\frac{1}{2}x+\frac{1}{4}\sin{2x}+C$
We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral.
\begin{equation*}
\frac{d}{dx}\left(\frac{1}{2}x+\frac{1}{4}\sin{2x}+C\right)=\frac{d}{dx}\left(\frac{1}{2}x\right)+\frac{d}{dx}\left(\frac{1}{4}\sin{2x}\right)+\frac{d}{dx}\left(C\right)
\end{equation*}
\begin{equation*}
=\left(\frac{1}{2}\right)+\left(\frac{1}{4}\cos{2x}\times2\right)+\left(0\right)
\end{equation*}
\begin{equation*}
=\frac{1}{2}+\frac{1}{2}\cos{2x}=\frac{1+\cos{2x}}{2}=\cos^2{x}
\end{equation*}
as desired.
