MCQ
If $n$ is any integer, then $\int_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^3}(2n + 1)x\,dx = } $
- A$x$
- B$1$
- ✓$0$
- DNone of these
✓
Answer
Correct option: C.
$0$
c
(c) Since $\cos (2n + 1)(\pi - x) = \cos [(2n + 1)\pi $$ - (2n + 1)x]$
(c) Since $\cos (2n + 1)(\pi - x) = \cos [(2n + 1)\pi $$ - (2n + 1)x]$
$= - \cos (2n + 1)x$ and ${\cos ^2}(\pi - x) = {\cos ^2}x$
So that $f(2a - x) = - f(x)$, and
hence by the property of definite integral
$\int_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^3}(2n + 1)x\,dx = 0} $.
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