int_{-pi}^pi frac{cos ^{2022} x}{1+(2022)^x} | vedclass

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Question

(A) Let $I = \int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x$  ...$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get:$I

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$\frac{2022 !}{2^{2022}((1011) !)^2} \pi$

Vedclass · Answer

(A) Let $I = \int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x$  ...$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get:$I = \int_{-\pi}^\pi \frac{\cos ^{2022}(-x)}{1+(2022)^{-x}} d x = \int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+\frac{1}{(2022)^x}} d x = \int_{-\pi}^\pi \frac{(2022)^x \cos ^{2022} x}{(2022)^x+1} d x$  ...(ii)Adding $(i)$ and (ii):$2I = \int_{-\pi}^\pi \frac{\cos ^{2022} x (1+(2022)^x)}{1+(2022)^x} d x = \int_{-\pi}^\pi \cos ^{2022} x d x$Since $\cos ^{2022} x$ is an even function,$2I = 2 \int_0^\pi \cos ^{2022} x d x$,so $I = \int_0^\pi \cos ^{2022} x d x = 2 \int_0^{\pi/2} \cos ^{2022} x d x$.Using Wallis' formula $\int_0^{\pi/2} \cos^n x d x = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:$I = 2 \cdot \frac{2021!!}{2022!!} \cdot \frac{\pi}{2} = \frac{2021!!}{2022!!} \pi = \frac{2021!! \cdot 2022!!}{(2022!!)^2} \pi = \frac{2022!}{(2^{1011} \cdot 1011!)^2} \pi = \frac{2022!}{2^{2022} (1011!)^2} \pi$.

$\int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x=$

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