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 $n$.