Question

Find the solution of the integral \[\int_{ - \pi }^\pi {{{\left( {\cos ax - \sin bx} \right)}^2}dx} \] where \[a\] and \[b\] are integers. Choose the correct option.
a. \[ - \pi \]
b. 0
c. \[\pi \]
d. \[2\pi \]

Answer 1

Hint—We will use the property \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and then integrate each value separately and applying the limits also from \[ - \pi \] to \[\pi \]. The value of the integral \[2\cos ax\sin bx = 0\] as \[\cos ax\sin bx\] is an odd function in the interval \[\left[ { - \pi ,\pi } \right]\].

Complete step-by-step solution
Consider the given integral,
\[I = \int_{ - \pi }^\pi {{{\left( {\cos ax - \sin bx} \right)}^2}dx} \]
Use the property \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] to expand the given function,
We get,
\[
  I = \int_{ - \pi }^\pi {\left( {{{\cos }^2}ax + {{\sin }^2}bx - 2\cos ax\sin bx} \right)dx} \\
   = \int_{ - \pi }^\pi {{{\cos }^2}axdx} + \int_{ - \pi }^\pi {{{\sin }^2}bxdx} - \int_{ - \pi }^\pi {2\cos ax\sin bxdx} \\
\]
Here, we know that \[\cos ax\sin bx\] is an odd function in the interval \[\left[ { - \pi ,\pi } \right]\].
Thus, from this we get, \[2\cos ax\sin bx = 0\].
Hence, put the value in the integral,
\[I = \int_{ - \pi }^\pi {{{\cos }^2}axdx} + \int_{ - \pi }^\pi {{{\sin }^2}bxdx} - 0\]
Since, we know that \[{\cos ^2}ax\] and \[{\sin ^2}bx\] are even functions in the interval so, we will change the interval from \[\left[ { - \pi ,\pi } \right]\] to \[\left[ {0,\pi } \right]\] and multiply the integral by 2.
Thus, we get,
\[I = \int_0^\pi {2{{\cos }^2}axdx} + \int_0^\pi {2{{\sin }^2}bxdx} \]
Further, we know that, \[{\cos ^2}ax = \dfrac{{1 + \cos 2ax}}{2}\] and \[{\sin ^2}bx = \dfrac{{1 - \cos 2bx}}{2}\]
Thus, put the values in the integral to simply further.
Thus, we get,
\[
  I = \int_0^\pi {\left( {1 + \cos 2ax} \right)dx} + \int_0^\pi {\left( {1 - \cos 2bx} \right)dx} \\
   = \int_0^\pi {\left( {1 + \cos 2ax + 1 - \cos 2bx} \right)dx} \\
   = \int_0^\pi {\left( {2 + \cos 2ax - \cos 2bx} \right)dx} \\
 \]
Now, we will integrate each part individually in the integral.
Thus,
\[
  I = 2\left[ x \right]_0^\pi + \left[ {\dfrac{{\sin 2ax}}{{2a}}} \right]_0^\pi - \left[ {\dfrac{{\sin 2bx}}{{2b}}} \right]_0^\pi \\
   = 2\left( {\pi - 0} \right) + \left( {\dfrac{{\sin 2a\pi }}{{2a}} - \dfrac{{\sin 2a\left( 0 \right)}}{{2a}}} \right) - \left( {\dfrac{{\sin 2b\pi }}{{2b}} - \dfrac{{\sin 2b\left( 0 \right)}}{{2b}}} \right) \\
   = 2\pi + \dfrac{{\sin 2a\pi }}{{2a}} - \dfrac{{\sin 2b\pi }}{{2b}} \\
\]
Now, as we know that \[\sin n\pi = 0\]
Here, \[\sin 2a\pi = 0\] and \[\sin 2b\pi = 0\] as \[a\] and \[b\] are integers.
Thus, we get that,
\[I = 2\pi \]
The option (d) is the correct option, as the solution of the integral is \[I = 2\pi \].

Note: The properties \[{\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}\] and \[{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\] makes the calculation of the integral easier. We must know that the function \[{\cos ^2}x\] and \[{\sin ^2}x\] are even functions and the function \[\cos x\sin x\] forms an odd function. Use the fact that \[\sin n\pi = 0\] for all the values of \[n\].