Answer to Question #136209 in Calculus for qwerty

Question

\int \:cos5xcos2xdx

Answer 1

$\textsf{Let}\hspace{0.1cm}y = \cos(5x)\cos(2x)\\ \textsf{It is known that}\\ \cos(A - B ) = \cos(A)\cos(B) + \sin(A)\sin(B)\hspace{0.1cm}\& \\ \cos(A + B ) = \cos(A)\cos(B) - \sin(A)\sin(B) \\ \textsf{Adding both equations yields,} \displaystyle\frac{\cos(A - B )}{2} + \frac{\cos(A + B)}{2}= \cos(A)\cos(B) \\ \therefore y = \frac{\cos(7x )}{2} + \frac{\cos(3x)}{2}\\ \begin{aligned} \int y\hspace{0.1cm}\mathrm{d}x & = \int \frac{\cos(7x )}{2} + \frac{\cos(3x)}{2} \hspace{0.1cm}\mathrm{d}x \\&= \frac{\sin(7x )}{14} + \frac{\sin(3x)}{6} + C \end{aligned}\\ \int \cos(5x)\cos(2x) \displaystyle\hspace{0.1cm}\mathrm{d}x = \frac{\sin(7x )}{14} + \frac{\sin(3x)}{6} + C\\ \textsf{Where}\hspace{0.1cm} C\hspace{0.1cm}\textsf{is an arbitrary constant}$