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Question: Prove that \[\int_0^{\dfrac{\pi }{2}} {{{\cos }^n}x\cos nx dx = \dfrac{1}{{{2^{\left( {n + 1} \right...

Prove that ∫0π2cos⁡nxcos⁡nxdx=12(n+1),∀n∈N\int_0^{\dfrac{\pi }{2}} {{{\cos }^n}x\cos nx dx = \dfrac{1}{{{2^{\left( {n + 1} \right)}}}},\forall n \in {\bf{N}}}

Explanation

Solution

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⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B
  2. Exponential formula: ax.ay=ax+y{a^x}.{a^y} = {a^{x + y}}
  3. Integration by Parts: ∫udv=uv−∫vdu\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 ………………………………(1)\left( 1 \right)
Then, by the sum of angles identity, we get
⇒In−1=∫0π2cos⁡n−1xcos⁡((n−1)x)dx\Rightarrow {I_{n - 1}} = \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\cos ((n - 1)x)dx}
⇒In−1=∫0π2cos⁡n−1xcos⁡(nx−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⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B, we get
⇒In−1=∫0π2cos⁡n−1x(cos⁡(nx)cos⁡x+sin⁡(nx)sin⁡x)dx\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
⇒In−1=∫0π2cosn−1xcos⁡(nx)cos⁡xdx+∫0π2cos⁡n−1xsin⁡(nx)sin⁡x)dx\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 ax⋅ay=ax+y{a^x} \cdot {a^y} = {a^{x + y}}, we get
⇒In−1=∫0π2cos⁡n−1+1xcos⁡(nx)dx+∫0π2cos⁡n−1xsin⁡(nx)sin⁡x)dx\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} }
⇒In−1=∫0π2cos⁡nxcos⁡(nx)dx+∫0π2cos⁡n−1xsin⁡(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
⇒In−1=In+∫0π2cos⁡n−1xsin⁡(nx)sin⁡x)dx\Rightarrow {I_{n - 1}} = {I_n} + \int_0^{\dfrac{\pi }{2}} {{{\cos }^{n - 1}}x\sin (nx)\sin x)dx}
⇒In−1=In+∫0π2sin⁡(nx)cos⁡n−1xsin⁡xdx\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)u = \sin (nx)and dv=cos⁡n−1xsin⁡xdxdv = {\cos ^{n - 1}}x\sin xdx

Now, we will differentiate the term uu. Therefore, we get

dudx=ncos⁡(nx) ⇒du=ncos⁡(nx)dx\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−1xsin⁡xdxdv = {\cos ^{n - 1}}x\sin xdx. Therefore, we get
v=−1ncos⁡nxv = \dfrac{{ - 1}}{n}{\cos ^n}x
Now, by using the integration by parts ∫udv=uv−∫vdu\int {udv = uv - \int {vdu} } , we get
In−1=In+[−1nsin⁡(nx)cos⁡nx]0π2+∫0π2cos⁡nxcos⁡(nx)dx{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
In−1=In+(0+0)+In{I_{n - 1}} = {I_n} + (0 + 0) + {I_n}
Adding the terms, we get
In−1=2In{I_{n - 1}} = 2{I_n}
By using this relation, we find that
In=In−12=In−222=⋯=I02n.{I_n} = \dfrac{{{I_{n - 1}}}}{2} = \dfrac{{{I_{n - 2}}}}{{{2^2}}} = \cdots = \dfrac{{{I_0}}}{{{2^n}}}.
Since,n∈Nn \in {\bf{N}}, we have n=1,2,3,......nn = 1,2,3,......n
Substituting n=1n = 1 in the equation In−1=2In{I_{n - 1}} = 2{I_n}, we get
⇒I1=I1−12=I02\Rightarrow {I_1} = \dfrac{{{I_{1 - 1}}}}{2} = \dfrac{{{I_0}}}{2}
⇒I1=π2.2\Rightarrow {I_1} = \dfrac{\pi }{{2.2}}
⇒I1=π4\Rightarrow {I_1} = \dfrac{\pi }{4}
We conclude that In=π2n+1{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.