# Evaluate the definite integral. {eq}\displaystyle \int_{0}^{\frac{\pi}{4}} \cos 2 x \sin (\sin 2 x)\ dx {/eq}.

## Question:

Evaluate the definite integral.

{eq}\displaystyle \int_{0}^{\frac{\pi}{4}} \cos 2 x \sin (\sin 2 x)\ dx {/eq}.

## Definite Integral:

The procedure for solving a definite integral is to first calculate the indefinite integral by applying the most convenient integration method to simplify and convert the integral into a common or standard integral. Finally, we compute the boundaries of the definite integral.

{eq}\eqalign{ & {\text{We're going to evaluate the definite integral }}\,\int_0^{\frac{\pi }{4}} {\cos 2x\sin \left( {\sin 2x} \right)dx} . \cr & {\text{First we must solve the indefinite integral }}\,\int {\cos 2x\sin \left( {\sin 2x} \right)d} x{\text{:}} \cr & {\text{Applying }}u{\text{ - substitution}}\,{\text{ }}\,u = \sin \left( {2x} \right){\text{:}} \cr & \,\,\,u = \sin 2x\,\,\,\, \Rightarrow du = 2\cos 2xdx\,\,\,\,\, \Rightarrow \cos 2xdx = \frac{{du}}{2} \cr & \,\,\,\,\, \Rightarrow \int {\cos 2x\sin \left( {\sin 2x} \right)dx} = \int {\sin \left( {\sin 2x} \right)} \cos 2xdx = \int {\sin u\,\frac{{du}}{2}} = \frac{1}{2}\int {\sin udu} \cr & {\text{using the standard integral }}\,\int {\sin vdv = - \cos v{\text{:}}} \cr & \,\,\,\,\,\,\frac{1}{2}\int {\sin udu} = - \frac{1}{2}\cos u \cr & {\text{undo }}u{\text{ - substitution }}\,u = \sin \left( {2x} \right){\text{ and adding a constant }}C{\text{:}} \cr & \,\,\,\,\,\,\int {\cos 2x\sin \left( {\sin 2x} \right)d} x = - \frac{1}{2}\cos \left( {\sin 2x} \right) + C \cr & {\text{thus:}} \cr & \,\,\,\,\,\,\,\int_0^{\frac{\pi }{4}} {\cos 2x\sin \left( {\sin 2x} \right)dx} = \left[ { - \frac{1}{2}\cos \left( {\sin 2x} \right)} \right]_{x = 0}^{x = \frac{\pi }{4}} \cr & {\text{calculating the upper and lower limits we have}}: \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ { - \frac{1}{2}\cos \left( {\sin 2\left( {\frac{\pi }{4}} \right)} \right)} \right] - \left[ { - \frac{1}{2}\cos \left( {\sin 2\left( 0 \right)} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ { - \frac{1}{2}\cos \left( {\sin \left( {\frac{\pi }{2}} \right)} \right)} \right] - \left[ { - \frac{1}{2}\cos \left( {\sin \left( 0 \right)} \right)} \right] \cr & {\text{simplifying:}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ { - \frac{1}{2}\cos \left( 1 \right)} \right] - \left[ { - \frac{1}{2}\cos \left( 0 \right)} \right] = - \frac{1}{2}\cos \left( 1 \right) + \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {1 - \cos \left( 1 \right)} \right) \cr & {\text{Therefore:}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\boxed{\int_0^{\frac{\pi }{4}} {\cos 2x\sin \left( {\sin 2x} \right)dx} = \frac{1}{2}\left( {1 - \cos \left( 1 \right)} \right)} \cr} {/eq}