Question
For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-
A.
$$\pi $$
B.
$$1$$
C.
$$0$$
D.
none of these
Answer :
$$0$$
Solution :
$$\eqalign{
& I = \int_0^\pi {{e^{{{\cos }^2}x\,}}\,} {\cos ^3}\left( {2n + 1} \right)x\,dx,\,n \in \,Z.....(1) \cr
& = \int_0^\pi {{e^{{{\cos }^2}\left( {\pi - x} \right)}}} \,{\cos ^3}\left[ {\left( {2n + 1} \right)\left( {\pi - x} \right)} \right]dx \cr
& {\text{Using }}\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \cr
& \therefore I = \int_0^\pi {{e^{{{\cos }^2}x}}} \,{\cos ^3}\left[ {\left( {2n + 1} \right)\pi - \left( {2n + 1} \right)x} \right]dx \cr
& I = \int_0^\pi {\left( { - {e^{{{\cos }^2}x}}\,{{\cos }^3}} \right)\left( {2n + 1} \right)x\,dx.....(2)} \cr
& {\text{Adding (1) and (2) we get}} \cr
& 2I = 0\,\,\,\, \Rightarrow I = 0 \cr} $$