Answer
\[ = \sin \,\left( {4{x^2} + 3} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {8x\cos \,\left( {4{x^2} + 3} \right)\,dx} \hfill \\
\hfill \\
set\,\,\,u = 4{x^2} + 3\,\,\,\,\,then\,\,\,\,du = 8xdx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int_{}^{} {\cos \,\left( u \right)du} \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
= \sin \,\left( u \right) + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = 4{x^2} + 3 \hfill \\
\hfill \\
= \sin \,\left( {4{x^2} + 3} \right) + C \hfill \\
\end{gathered} \]
