   Chapter 7.2, Problem 60E

Chapter
Section
Textbook Problem

# Use a graph of the integrated to guess the value of the integral. Then use the methods of this section to prove that your guess is correct. ∫ 0 2 sin 2 π x   cos 5 π x   d x

To determine

To evaluate:

the value of integral from the graph of the integrand and then perform the integration to verify.

Explanation

Trigonometric integral of the form sinmxcosnxdx can be solved using the identity for sinAcosA.

Given:

02sin2πxcos5πxdx.

Formula used:

The identity sinAcosB=12[sin(AB)+sin(A+B)].

Calculation:

The graph of integrand from x=0 to 2 is as shown below:

Notice from the graph that the shape of the curve from x=0 to 1 is repeated for x=1 to 2, but inverted about the x-axis. Hence for each bell shapes above the x-axis, there exists an equal bell shape below the x-axis. So, the total area above x-axis under the graph is equal to the area under the graph below the x-axis. Hence, the integral from 0 to 2 will be zero. Now, perform the integration and confirm is it evaluates to zero.

Use the identity sinAcosB=12[sin(AB)+sin(A+B)] with A as 2πx and B as 5πx:

02sin2πxcos5πxdx=1202(sin(2πx5πx)+sin(2πx+5πx))dx=

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