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Question

Question: What is the integral of \[\int {\dfrac{{\cos x}}{x}} dx\]?...

What is the integral of cosxxdx\int {\dfrac{{\cos x}}{x}} dx?

Explanation

Solution

Hint : To find the integral of cosxx\dfrac{{\cos x}}{x}, we will use the Taylor series expansion of cosx\cos x to expand. Then we will divide by xx to find the expansion of cosxx\dfrac{{\cos x}}{x}. And finally, we will integrate this series term by term to give a power series expansion for the integral of cosxx\dfrac{{\cos x}}{x}.

Complete step-by-step answer :
We have to find the integral of cosxx\dfrac{{\cos x}}{x}.
We can’t do this integration in terms of elementary functions. We can do it in terms of infinite series using Taylor series expansion of cosx\cos x.
We know that the Taylor series expansion of cosx\cos x is
cosx=1x22!+x44!x66!+...\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + ...
We require the integral of cosxx\dfrac{{\cos x}}{x}. So, we will divide the Taylor series expansion of cosx\cos x by xx.
Dividing Taylor series expansion of cosx\cos x by xx gives us an infinite series expansion of cosxx\dfrac{{\cos x}}{x}. Therefore, we get the expansion as
cosxx=1xx2!+x34!x56!+...\Rightarrow \dfrac{{\cos x}}{x} = \dfrac{1}{x} - \dfrac{x}{{2!}} + \dfrac{{{x^3}}}{{4!}} - \dfrac{{{x^5}}}{{6!}} + ...
Now, we will integrate this series term by term which will give a power series expansion for the integral of cosxx\dfrac{{\cos x}}{x}. Therefore, we get
cosxx=1xx2!+x34!x56!+...\Rightarrow \int {\dfrac{{\cos x}}{x}} = \int {\dfrac{1}{x}} - \int {\dfrac{x}{{2!}}} + \int {\dfrac{{{x^3}}}{{4!}}} - \int {\dfrac{{{x^5}}}{{6!}}} + ...
Integrating right hand side of the above equation using the formula of integration, we get
cosxx=lnxx22×2!+x44×4!x66×6!+...+C\Rightarrow \int {\dfrac{{\cos x}}{x}} = \ln x - \dfrac{{{x^2}}}{{2 \times 2!}} + \dfrac{{{x^4}}}{{4 \times 4!}} - \dfrac{{{x^6}}}{{6 \times 6!}} + ... + C
Where, CC is the constant of integration.
Therefore, integral of cosxx\dfrac{{\cos x}}{x} is lnxx22×2!+x44×4!x66×6!+...+C\ln x - \dfrac{{{x^2}}}{{2 \times 2!}} + \dfrac{{{x^4}}}{{4 \times 4!}} - \dfrac{{{x^6}}}{{6 \times 6!}} + ... + C.
So, the correct answer is “lnxx22×2!+x44×4!x66×6!+...+C\ln x - \dfrac{{{x^2}}}{{2 \times 2!}} + \dfrac{{{x^4}}}{{4 \times 4!}} - \dfrac{{{x^6}}}{{6 \times 6!}} + ... + C”.

Note : One important point to note here is that this is one of those integrals that can’t be done in terms of elementary functions. We can do it in terms of infinite series and we can use various numerical methods to do the definite integration. Also, this sum approaches zero so that the definite integral is lnx\ln x up to an integration constant.