How do you integrate $\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} $, $m,n \in Z$?

Hint: In this question we have to find the integral value of the given function, we will find by using the trigonometric identity, $\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = 2\sin a\cos b$, then we will distribute the integration and by using the integration formula $\int {\sin x} dx = - \cos x$, then we will apply the intervals , then we will get the required result.

Complete step by step solution:
Given function is $\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} $,
We know that $\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = 2\sin a\cos b$, using the formula we get,
$ \Rightarrow \sin \left( {mx} \right)\cos \left( {nx} \right) = \dfrac{1}{2}\left[ {\sin \left( {mx + nx} \right) + \sin \left( {mx - nx} \right)} \right]$,
Now substituting the value, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = \int\limits_0^{2\pi } {\dfrac{1}{2}\left[ {\sin \left( {mx + nx} \right) + \sin \left( {mx - nx} \right)} \right]dx} $,
Now taking out the constant we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = \dfrac{1}{2}\int\limits_0^{2\pi } {\left[ {\sin \left( {m + n} \right)x + \sin \left( {m - n} \right)x} \right]dx} $,
Now distribute the integration to all terms, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = \dfrac{1}{2}\int\limits_0^{2\pi } {\sin \left( {m + n} \right)x} dx + \dfrac{1}{2}\int\limits_0^{2\pi } {\sin \left( {m - n} \right)x} dx$,
Now applying the integration we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {m + n} \right)x}}{{m + n}}} \right]_{_0}^{2\pi } - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {m - n} \right)x}}{{m - n}}} \right]_{_0}^{2\pi } + C$,
Now applying intervals we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {m + n} \right)\left( {2\pi } \right)}}{{m + n}} - \dfrac{{\cos \left( {m + n} \right)0}}{{m + n}}} \right] - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {m - n} \right)\left( {2\pi } \right)}}{{m - n}} - \dfrac{{\cos \left( {m - n} \right)0}}{{m - n}}} \right] + C$Now simplifying we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{m + n}} - \dfrac{{\cos 0}}{{m + n}}} \right] - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {\left( {m - n} \right)\left( {2\pi } \right)} \right)}}{{m - n}} - \dfrac{{\cos 0}}{{m - n}}} \right] + C$
Now we know that $\cos 0 = 1$, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{m + n}} - \dfrac{1}{{m + n}}} \right] - \dfrac{1}{2}\left[ {\dfrac{{\cos \left( {\left( {m - n} \right)\left( {2\pi } \right)} \right)}}{{m - n}} - \dfrac{1}{{m - n}}} \right] + C$
Now distributing the constant, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m + n} \right)}}} \right] + \dfrac{1}{{2\left( {m + n} \right)}} - \left[ {\dfrac{{\cos \left( {\left( {m - n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m - n} \right)}}} \right] + \dfrac{1}{{2\left( {m - n} \right)}}C$
Now simplifying we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m + n} \right)}}} \right] + - \left[ {\dfrac{{\cos \left( {\left( {m - n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m - n} \right)}}} \right] + \dfrac{1}{{2\left( {m + n} \right)}} + \dfrac{1}{{2\left( {m - n} \right)}}C$
Now taking L.C.M, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m + n} \right)\left( {m - n} \right)}} + \dfrac{{\left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {m + n} \right)\left( {m - n} \right)}}} \right]$$ + \dfrac{{2\left( {m - n} \right) + 2\left( {m + n} \right)}}{{4\left( {m + n} \right)\left( {m - n} \right)}} + C$,
Now simplifying we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$$ + \dfrac{{2m - 2n + 2m + 2n}}{{4\left( {{m^2} - {n^2}} \right)}} + C$,
Now simplifying we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$
$ + \dfrac{{4m}}{{4\left( {{m^2} - {n^2}} \right)}} + C$,
Now further simplification we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$$ + \dfrac{m}{{\left( {{m^2} - {n^2}} \right)}} + C$,
Now again simplifying the expression we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right)}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$$ + \dfrac{{2m}}{{2\left( {{m^2} - {n^2}} \right)}} + C$,
Now by adding as the denominators are equal, we get,
$ \Rightarrow \int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} = - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + 2m}}{{2\left( {{m^2} - {n^2}} \right)}}} \right] + C$
So, the integral value is $ - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + 2m}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$.

$\therefore $ The integral value of the given function $\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} $ will be equal to
$ - \left[ {\dfrac{{\left( {m - n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + \left( {m + n} \right)\cos \left( {\left( {m + n} \right)\left( {2\pi } \right)} \right) + 2m}}{{2\left( {{m^2} - {n^2}} \right)}}} \right]$.

Note: An indefinite is an integral that does not contain the upper and lower limit. Indefinite integral is also known as anti-derivative or Prime integral. Indefinite integral of a function f is generally a differentiable function F whose derivative is equal to the original function f. It is represented as,
$\int {f\left( x \right)dx} = F\left( x \right) + C$,
Some of the important formulas that we use while solving integration problems are given below:
$\int {dx = x + c} $,
$\int {adx = ax + c} $,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$,
$\int {\dfrac{1}{x}dx} = \ln x + c$,
$\int {{e^x}dx = {e^x} + c} $,
$\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c} $,
$\int {\sin xdx = - \cos x + c} $,
$\int {\cos xdx = \sin x + c} $,
$\int {{{\sec }^2}xdx = \tan x + c} $,
$\int {{{\csc }^2}xdx = - \cot x + C} $,
$\int {\sec x\tan xdx = \sec x + C} $,
$\int {\csc x\cot xdx = - \csc x + C} $,
$\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c} $,
$\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c} $.