_{-π}^{π} {²022} x}{1+(2022)^{x}} \, dx =

Correct Answer is : (2022)!22022((1011)!)2π\; {(2022)!}{2²022} ((1011)!)² π22022((1011)!)2(2022)!π

\; {(2022)!}{2²022} ((1011)!)² π

({}²022}C₁011}) \; {π}{2¹011}}

\; {(2022)!}{(1011)! 2²022}} π

Correct Answer is : (2022)!22022((1011)!)2π\; {(2022)!}{2²022} ((1011)!)² π22022((1011)!)2(2022)!π

Test Index

AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int\limits_{-\pi}^{\pi} \frac{\cos^{2022} x}{1+(2022)^{x}} \, dx =

Solution: 👈: Video Solution

I = \int_{-\pi_1}^{\pi} \frac{\cos^{2022}}{1 + (2022)^x} \, dx

\text{ ...(i)}

\Rightarrow I = \int_{-\pi_1}^{\pi} \frac{\cos^{2022}(-X)}{1 + (2022)^{-x}} \, dx

\text{...}\left[ \because \int_{a}^{b} f(a+b-x) \, dx = \int_{a}^{b} f(x) \, dx \right]

I = \int_{-\pi_1}^{\pi} \frac{\cos^{2022} x \cdot (2022)^x}{(2022)^x + 1} \, dx

\text{ ...(ii)}

On adding

\text{Eqs (i) and (ii)}

2I = \int_{-\pi_1}^{\pi} coss^{2022} x \, dx

\Rightarrow 2I = 2 \int_{0}^{\pi} \cos^{2022} x \, dx

\text{...[ }\because \cos x^{2022} \text{ is an even function]}

\therefore I = \int_{0}^{\pi} \cos^{2022} x \, dx = 2 \int_{0}^{\pi/2} \cos^{2022} x \, dx

2 \frac{\pi}{2^{2022} + 1} {}^{2022}C_{1011}

\text{...}[ \because \int_{0}^{\pi/2} \cos^{2n} x \, dx = \frac{\pi}{2^{2n} + 1} \cdot {}^{2n}C_n \text{ ]}

= \frac{\pi}{2^{2022}} \cdot \frac{2022!}{1011! (2022-1011)!} = \frac{\pi \cdot 2022!}{2^{2022} ((1011)!)^2}

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