Question

If \[{U_n} = \int\limits_0^\pi {\dfrac{{1 - \cos nx}}{{1 - \cos x}}} \] show that \[{U_{n + 2}} + {U_n} = 2{U_{n + 1}}\] .

Answer 1

Hint: Integration is the process of finding the antiderivative. To get the given equation, \[{U_{n + 2}} + {U_n} = 2{U_{n + 1}}\] ; we need to apply formulas based on the given integral function such that by expanding the terms of respective formula we need to integrate the terms directly.
Formula used:
 \[\sin \left( a \right) - \sin \left( b \right) = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)\]
 \[\cos \left( a \right) - \cos \left( b \right) = - 2\sin \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)\]

Complete step-by-step answer:
Let us write the given equation:
 \[{U_n} = \int\limits_0^\pi {\dfrac{{1 - \cos nx}}{{1 - \cos x}}} \]
And we need to show that:
 \[{U_{n + 2}} + {U_n} = 2{U_{n + 1}}\]
As, from the given equation we have,
 \[{U_{n + 2}} - {U_{n + 1}}\]
Substitute the value of \[{U_{n + 2}}\] and \[{U_{n + 1}}\] from the given function of \[{U_n}\] i.e.,
 \[{U_n} = \int\limits_0^\pi {\dfrac{{1 - \cos nx}}{{1 - \cos x}}} \] we get:
= \[\int\limits_0^\pi {\left( {\dfrac{{1 - \cos \left( {n + 2} \right)x}}{{1 - \cos x}}} \right)} dx - \int\limits_0^\pi {\left( {\dfrac{{1 - \cos \left( {n + 1} \right)x}}{{1 - \cos x}}} \right)} dx\]
= \[\int\limits_0^\pi {\left( {\dfrac{{\cos \left( {n + 1} \right)x - \cos \left( {n + 2} \right)x}}{{1 - \cos x}}} \right)} dx\]
As, the obtained equation is of the form \[\cos \left( a \right) - \cos \left( b \right)\] , hence apply the formula of \[\cos \left( a \right) - \cos \left( b \right)\] i.e.,
 \[\cos \left( a \right) - \cos \left( b \right) = - 2\sin \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)\]
= \[\int\limits_0^\pi {\left( {\dfrac{{ - 2\sin \left( {\dfrac{{2n + 3}}{2}} \right)x \cdot \sin \left( {\dfrac{{n + 1 - n - 2}}{2}} \right)x}}{{2{{\sin }^2}\dfrac{x}{2}}}} \right)} dx\]
= \[\int\limits_0^\pi {\left( {\dfrac{{\sin \left( {n + \dfrac{3}{2}} \right)x \cdot \sin \dfrac{x}{2}}}{{{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}} \right)} dx\]
There is a common term involved of \[\sin \left( {\dfrac{x}{2}} \right)\] , hence we get:
= \[\int\limits_0^\pi {\left( {\dfrac{{\sin \left( {n + \dfrac{3}{2}} \right)x}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right)} dx\]
Hence, we get the value of \[{U_{n + 2}} - {U_{n + 1}}\] as:
= \[\int\limits_0^\pi {\left( {\dfrac{{\sin \left( {n + \dfrac{3}{2}} \right)x}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right)} dx\] ………………… 1
Hence, for \[{U_{n + 1}} - {U_n}\] we get:
= \[\int\limits_0^\pi {\left( {\dfrac{{\sin \left( {n + \dfrac{1}{2}} \right)x}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right)} dx\] ……………….. 2
Now, subtract equation 2 from equation 1 we get:
 \[\left( {{U_{n + 2}} - {U_{n + 1}}} \right) - \left( {{U_{n + 1}} - {U_n}} \right) = {U_{n + 2}} - 2{U_{n + 1}} + {U_n}\]
 \[ \Rightarrow \] \[{U_{n + 2}} - 2{U_{n + 1}} + {U_n}\] = \[\int\limits_0^\pi {\left( {\dfrac{{\sin \left( {n + \dfrac{3}{2}} \right)x - \sin \left( {n + \dfrac{1}{2}} \right)x}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right)} dx\]
As, the obtained equation is of the form \[\sin \left( a \right) - \sin \left( b \right)\] , hence apply the formula of \[\sin \left( a \right) - \sin \left( b \right)\] i.e.,
 \[\sin \left( a \right) - \sin \left( b \right) = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)\]
= \[\int\limits_0^\pi {\left( {\dfrac{{2\cos \left( {n + 1} \right)x .\sin \left( {\dfrac{x}{2}} \right)}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right)} dx\]
Hence, applying the formula we get:
= \[2\int\limits_0^\pi {\left( {\cos \left( {n + 1} \right)x} \right)} dx\]
Now, let us apply the integration directly:
= \[2\left( {\dfrac{{\sin \left( {n + 1} \right)x}}{{n + 1}}} \right] _0^\pi \]
Simplifying the terms, we get
 \[{U_{n + 2}} - 2{U_{n + 1}} + {U_n} = 0\]
Therefore,
 \[ \Rightarrow \] \[{U_{n + 2}} + {U_n} = 2{U_{n + 1}}\]

Note: In the integration of a function, if the integrand involves any kind of trigonometric function, then we use trigonometric identities to simplify the function that can be easily integrated. The integration of these standard integrands can be easily found using a direct form of integration method. There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. Hence, based on the function given we can find the integration of the equation.