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Question: How do you integrate \(\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} \)...

How do you integrate ∫02πsin⁡(mx)cos⁡(nx)dx\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} , m,n∈Zm,n \in Z?

Explanation

Solution

In this question we have to find the integral value of the given function, we will find by using the trigonometric identity, sin⁡(a+b)+sin⁡(a−b)=2sin⁡acos⁡b\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 ∫sin⁡xdx=−cos⁡x\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 ∫02πsin⁡(mx)cos⁡(nx)dx\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} ,
We know that sin⁡(a+b)+sin⁡(a−b)=2sin⁡acos⁡b\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = 2\sin a\cos b, using the formula we get,
⇒sin⁡(mx)cos⁡(nx)=12[sin⁡(mx+nx)+sin⁡(mx−nx)]\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=∫02π12[sin⁡(mx+nx)+sin⁡(mx−nx)]dx\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=12∫02π[sin⁡(m+n)x+sin⁡(m−n)x]dx\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=12∫02πsin⁡(m+n)xdx+12∫02πsin⁡(m−n)xdx\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−12[cos⁡(m+n)xm+n]02π−12[cos⁡(m−n)xm−n]02π+C\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−12[cos⁡(m+n)(2π)m+n−cos⁡(m+n)0m+n]−12[cos⁡(m−n)(2π)m−n−cos⁡(m−n)0m−n]+C\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] + CNow simplifying we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−12[cos⁡((m+n)(2π))m+n−cos⁡0m+n]−12[cos⁡((m−n)(2π))m−n−cos⁡0m−n]+C\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\cos 0 = 1, we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−12[cos⁡((m+n)(2π))m+n−1m+n]−12[cos⁡((m−n)(2π))m−n−1m−n]+C\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[cos⁡((m+n)(2π))2(m+n)]+12(m+n)−[cos⁡((m−n)(2π))2(m−n)]+12(m−n)C\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[cos⁡((m+n)(2π))2(m+n)]+−[cos⁡((m−n)(2π))2(m−n)]+12(m+n)+12(m−n)C\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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))2(m+n)(m−n)+(m+n)cos⁡((m+n)(2π))2(m+n)(m−n)]\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] +2(m−n)+2(m+n)4(m+n)(m−n)+C + \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,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))2(m2−n2)]\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] +2m−2n+2m+2n4(m2−n2)+C + \dfrac{{2m - 2n + 2m + 2n}}{{4\left( {{m^2} - {n^2}} \right)}} + C,
Now simplifying we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))2(m2−n2)]\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]
+4m4(m2−n2)+C+ \dfrac{{4m}}{{4\left( {{m^2} - {n^2}} \right)}} + C,
Now further simplification we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))2(m2−n2)]\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] +m(m2−n2)+C + \dfrac{m}{{\left( {{m^2} - {n^2}} \right)}} + C,
Now again simplifying the expression we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))2(m2−n2)]\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] +2m2(m2−n2)+C + \dfrac{{2m}}{{2\left( {{m^2} - {n^2}} \right)}} + C,
Now by adding as the denominators are equal, we get,
⇒∫02πsin⁡(mx)cos⁡(nx)dx=−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))+2m2(m2−n2)]+C\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 −[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))+2m2(m2−n2)] - \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 ∫02πsin⁡(mx)cos⁡(nx)dx\int\limits_0^{2\pi } {\sin \left( {mx} \right)\cos \left( {nx} \right)dx} will be equal to
−[(m−n)cos⁡((m+n)(2π))+(m+n)cos⁡((m+n)(2π))+2m2(m2−n2)]- \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,
∫f(x)dx=F(x)+C\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:
∫dx=x+c\int {dx = x + c} ,
∫adx=ax+c\int {adx = ax + c} ,
∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c,
∫1xdx=ln⁡x+c\int {\dfrac{1}{x}dx} = \ln x + c,
∫exdx=ex+c\int {{e^x}dx = {e^x} + c} ,
∫axdx=axln⁡a+c\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c} ,
∫sin⁡xdx=−cos⁡x+c\int {\sin xdx = - \cos x + c} ,
∫cos⁡xdx=sin⁡x+c\int {\cos xdx = \sin x + c} ,
∫sec⁡2xdx=tan⁡x+c\int {{{\sec }^2}xdx = \tan x + c} ,
∫csc⁡2xdx=−cot⁡x+C\int {{{\csc }^2}xdx = - \cot x + C} ,
∫sec⁡xtan⁡xdx=sec⁡x+C\int {\sec x\tan xdx = \sec x + C} ,
∫csc⁡xcot⁡xdx=−csc⁡x+C\int {\csc x\cot xdx = - \csc x + C} ,
∫11−x2dx=sin⁡−1x+c\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c} ,
∫11+x2dx=tan⁡−1x+c\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c} .