Correct Answer - Option 2 :
\( - \frac{4}{{{n^2}}}\) when n is odd and 0 when n is even
Concept:
Property of definite integral:
If f(x) is an odd function then:
\(\int\limits_{ - a}^a {f(x)dx} = 0\)
If f(x) is an even function then:
\(\int\limits_{ - a}^a {f(x)dx} =2\int\limits_0^a {f(x)dx}\)
Calculation:
\(\mathop \smallint \limits_{ - π }^π \left| x \right|\cos nxdx\)
The given function f(x) = |x| cos nx is even function. So, the given integral can be reduced to:
\( = 2\mathop \smallint \limits_0^π \left| x \right|\cos nxdx\)
\( = 2\mathop \smallint \limits_0^π x\cos nxdx\)
\(\smallint x\cos nxdx = x\smallint \cos nxdx - \smallint \frac{{\sin nx}}{n}dx\)
\( = x\left( {\frac{{\sin nx}}{n}} \right) + \frac{1}{{{n^2}}}\left( {\cos nx} \right)\)
\( = 2\left[ {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}} \right]_0^π \)
\( = 2\left[ {\frac{{\cos nπ - 1}}{{{n^2}}}} \right]\)
\( = -\frac{2}{n^2}(1-cos nπ)\)
Now cos π = -1, cos 2π = 1, cos 3π = -1, and so on..
So, in general, we can write, cos nπ = (-1)n;
\( = \frac{{ - 2}}{{{n^2}}}\left( {1 - {{\left( { - 1} \right)}^n}} \right)\)
= 0 when n is even
\( = - \frac{4}{{{n^2}}}\) when n is odd
