Question

Evaluate \[ \int_{-\pi}^{\pi}\frac{\cos^{2022}x}{1+(2022)^x}\,dx \] =

• A. \(\dfrac{(2022)!}{2^{2022}\left((1011)!\right)^2}\pi\)
• B. \({}^{2022}C_{1011}\pi\)
• C. \({}^{2022}C_{1011}\dfrac{\pi}{2^{1011}}\)
• D. \(\dfrac{(2022)!}{\left((1011)!\right)^2 2^{2022}}\pi\)

Hint:

For integrals over symmetric limits, if the denominator has terms like \(1+a^x\), use the substitution \(x\to -x\) and add both forms of the integral.

Answer 1

The Correct Option is A

Step 1: Let the integral be \(I\).
Let \[ I=\int_{-\pi}^{\pi}\frac{\cos^{2022}x}{1+(2022)^x}\,dx \]

Step 2: Use the substitution \(x\to -x\).
Since the limits are symmetric, substitute \(x\to -x\).
Then, \[ I=\int_{-\pi}^{\pi}\frac{\cos^{2022}(-x)}{1+(2022)^{-x}}\,dx \] Since cosine is an even function, \[ \cos(-x)=\cos x \] So, \[ \cos^{2022}(-x)=\cos^{2022}x \] Therefore, \[ I=\int_{-\pi}^{\pi}\frac{\cos^{2022}x}{1+(2022)^{-x}}\,dx \]

Step 3: Simplify the transformed denominator.
Now, \[ \frac{1}{1+(2022)^{-x}} = \frac{(2022)^x}{1+(2022)^x} \] Hence, \[ I=\int_{-\pi}^{\pi}\frac{(2022)^x\cos^{2022}x}{1+(2022)^x}\,dx \]

Step 4: Add both forms of \(I\).
Original form: \[ I=\int_{-\pi}^{\pi}\frac{\cos^{2022}x}{1+(2022)^x}\,dx \] Transformed form: \[ I=\int_{-\pi}^{\pi}\frac{(2022)^x\cos^{2022}x}{1+(2022)^x}\,dx \] Adding, \[ 2I=\int_{-\pi}^{\pi} \frac{\cos^{2022}x+(2022)^x\cos^{2022}x}{1+(2022)^x}\,dx \] \[ 2I=\int_{-\pi}^{\pi} \frac{\cos^{2022}x\left(1+(2022)^x\right)}{1+(2022)^x}\,dx \] Thus, \[ 2I=\int_{-\pi}^{\pi}\cos^{2022}x\,dx \] Hence, \[ I=\frac{1}{2}\int_{-\pi}^{\pi}\cos^{2022}x\,dx \]

Step 5: Use the standard even power cosine integral.
Since \(\cos^{2022}x\) is even, \[ \int_{-\pi}^{\pi}\cos^{2022}x\,dx = 2\int_{0}^{\pi}\cos^{2022}x\,dx \] Also, for \(n\in \mathbb{N}\), \[ \int_{-\pi}^{\pi}\cos^{2n}x\,dx = \frac{2\pi(2n)!}{2^{2n}(n!)^2} \] Here, \[ 2n=2022 \] So, \[ n=1011 \] Therefore, \[ \int_{-\pi}^{\pi}\cos^{2022}x\,dx = \frac{2\pi(2022)!}{2^{2022}\left((1011)!\right)^2} \]

Step 6: Substitute in \(I\).
Since, \[ I=\frac{1}{2}\int_{-\pi}^{\pi}\cos^{2022}x\,dx \] we get \[ I=\frac{1}{2}\cdot \frac{2\pi(2022)!}{2^{2022}\left((1011)!\right)^2} \] \[ I= \frac{(2022)!}{2^{2022}\left((1011)!\right)^2}\pi \]

Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{(2022)!}{2^{2022}\left((1011)!\right)^2}\pi} \]