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$\int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x=$
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Verified Answer
The correct answer is:
$\frac{(2022) !}{2^{2022}((1011) !)^2} \pi$
$I=\int_{-\pi}^\pi \frac{\cos ^{2022}}{1+(2022)^x} d x$ ...(i)
$\Rightarrow I=\int_{-\pi}^\pi \frac{\cos ^{2022} x \cdot\left(2022^x\right.}{\left(202 Z^x+1\right.} d x$ ...(ii)
On adding Eqs. (i) and (ii),
$2 I=\int_{-\pi}^\pi \cos ^{2022} x d x$
$\therefore \quad I=\int_0^\pi \cos ^{2022} x d x=2 \int_0^{\pi / 2} \cos ^{2022} x d x$
$=\frac{\pi}{2^{2022}} \cdot \frac{2022 !}{1011 !(2022-1011) !}=\frac{\pi \cdot 2022 !}{2^{2022}((1011) !)^2}$
$\Rightarrow I=\int_{-\pi}^\pi \frac{\cos ^{2022} x \cdot\left(2022^x\right.}{\left(202 Z^x+1\right.} d x$ ...(ii)
On adding Eqs. (i) and (ii),
$2 I=\int_{-\pi}^\pi \cos ^{2022} x d x$
$\therefore \quad I=\int_0^\pi \cos ^{2022} x d x=2 \int_0^{\pi / 2} \cos ^{2022} x d x$
$=\frac{\pi}{2^{2022}} \cdot \frac{2022 !}{1011 !(2022-1011) !}=\frac{\pi \cdot 2022 !}{2^{2022}((1011) !)^2}$
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