Question

Prove that \[\int_0^{\dfrac{\pi }{2}} {{{\cos }^n}x\cos nx dx = \dfrac{1}{{{2^{\left( {n + 1} \right)}}}},\forall n \in {\bf{N}}} \]

Answer 1

Hint:
Here, we have to prove the given equation. First, we will use the concept of integral, reduction formula and trigonometric formula. Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts.

Formula Used:
We will use the following formulas:
1) Trigonometric Formula: \[\cos (A - B) = \cos A\cos B + \sin A\sin B\]
2) Exponential formula: \[{a^x}.{a^y} = {a^{x + y}}\]
3) Integration by Parts: \[\int {udv = uv - \int {vdu} } \]

Complete step by step solution:
Let \[{I_n} = \mathop \smallint \nolimits_0^{\dfrac{\pi }{2}} {\cos ^n}x\cos nxdx\] ………………………………\[\left( 1 \right)\]
Then, by the sum of angles identity, we get
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\cos ((n - 1)x)dx} \]
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\cos (nx - x)dx} \]
Now, by using the trigonometric formula \[\cos (A - B) = \cos A\cos B + \sin A\sin B\], we get
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x(\cos (nx)\cos x + \sin (nx)\sin x)dx} \]
Multiplying the terms, we have
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {co{s^{n - 1}}x\cos (nx)\cos xdx + \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\sin (nx)\sin x)dx} } \]
By using the exponential formula \[{a^x} \cdot {a^y} = {a^{x + y}}\], we get
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1 + 1}}x\cos (nx)dx + \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\sin (nx)\sin x)dx} } \]
\[ \Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^n}x\cos (nx)dx + \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\sin (nx)\sin x)dx} } \]
From equation (1), we get
\[ \Rightarrow {I_{n - 1}} = {I_n} + \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\sin (nx)\sin x)dx} \]
\[ \Rightarrow {I_{n - 1}} = {I_n} + \int_0^{\dfrac{\pi }{2}} {\sin (nx){{\cos }^{n - 1}}x\sin xdx} \]
When doing the integration by parts the first function has to be differentiated and the second function has to be integrated.

Applying integration by parts with \[u = \sin (nx)\]and \[dv = {\cos ^{n - 1}}x\sin xdx\]

Now, we will differentiate the term \[u\]. Therefore, we get

\[\begin{array}{l}\dfrac{{du}}{{dx}} = n\cos (nx)\\ \Rightarrow du = n\cos (nx)dx\end{array}\]

Now, we will integrate the term \[dv = {\cos ^{n - 1}}x\sin xdx\]. Therefore, we get
\[v = \dfrac{{ - 1}}{n}{\cos ^n}x\]
Now, by using the integration by parts \[\int {udv = uv - \int {vdu} } \], we get
\[{I_{n - 1}} = {I_n} + \left[ { - \dfrac{1}{n}\sin (nx)\cos nx} \right]_0^{\dfrac{\pi }{2}} + \int_0^{\dfrac{\pi }{2}} {{{\cos }^n}x\cos (nx)dx} \]
Now, substituting the limits, we get
\[{I_{n - 1}} = {I_n} + (0 + 0) + {I_n}\]
Adding the terms, we get
\[{I_{n - 1}} = 2{I_n}\]
By using this relation, we find that
\[{I_n} = \dfrac{{{I_{n - 1}}}}{2} = \dfrac{{{I_{n - 2}}}}{{{2^2}}} = \cdots = \dfrac{{{I_0}}}{{{2^n}}}.\]
Since,\[n \in {\bf{N}}\], we have \[n = 1,2,3,......n\]
Substituting \[n = 1\] in the equation \[{I_{n - 1}} = 2{I_n}\], we get
\[ \Rightarrow {I_1} = \dfrac{{{I_{1 - 1}}}}{2} = \dfrac{{{I_0}}}{2}\]
\[ \Rightarrow {I_1} = \dfrac{\pi }{{2.2}}\]
\[ \Rightarrow {I_1} = \dfrac{\pi }{4}\]
We conclude that \[{I_n} = \dfrac{\pi }{{{2^{n + 1}}}}\]

Therefore, \[{I_n} = \mathop \smallint \nolimits_0^{\dfrac{\pi }{2}} {\cos ^n}x\cos nx dx = \dfrac{\pi }{{{2^{n + 1}}}}\]

Note:
We have integrated the function using integration by parts. Integration by parts is a special kind of integration method when two functions are multiplied together. This method is also termed as partial integration. The first function has to be identified according to the ILATE rule. ILATE stands for Inverse Trigonometric Function, Logarithmic Function, Algebraic Function, Trigonometric function, Exponential Function.