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.

Answer and Explanation: 1

{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}


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Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.


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