Question 18. If the value of the integral \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^{2}\cos x}{1+\pi^{x}}+\frac{1+\sin^{2}x}{1+e^{\sin x^{2023}}}\right)dx=\frac{\pi}{4}(\pi+a)-2, then the value of a is
(1) 3
(2) -\frac{3}{2}
(3) 2
(4) \frac{3}{2}
Answer (1)
Explanation:
Let
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^{2}\cos x}{1+\pi^{x}}+\frac{1+\sin^{2}x}{1+e^{\sin x^{2023}}}\right)dx.
Replace x by -x.
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^{2}\cos x}{1+\pi^{-x}}+\frac{1+\sin^{2}x}{1+e^{\sin(-x)^{2023}}}\right)dx.
Add the two expressions.
2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{2}\cos x+1+\sin^{2}x\right)dx.
Evaluate each term.
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^{2}\cos x\,dx=\frac{\pi^{2}}{2}.
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\,dx=\pi.
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{2}x\,dx=\frac{\pi}{2}.
So, 2I=\frac{\pi^{2}}{2}+\frac{3\pi}{2}-4.
Hence, I=\frac{\pi^{2}}{4}+\frac{3\pi}{4}-2.
Comparing with \frac{\pi}{4}(\pi+a)-2,
we get a=3.
Therefore, option (1) is correct.