If the value of the integral ∫−π2π2(x2cosx1+πx+1+sin2x1+esinx2023)dx=π4(π+a)−2, then the value of a is

I=∫−π/2π/2(x2cosx1+πx+1+sin2x1+esinx2023)dx I=∫−π/2π/2(x2cosx1+π−x+1+sin2x1+esin(−x)2023)dx On Adding, we get 2I=∫−π/2π/2(x2cosx+1+sin2x)dxOn solving I=π24+3π4−2 a=3

If the value of the integral ∫−π2π2(x2cosx1+πx+1+sin2x1+esinx2023)dx=π4(π+a)−2, then the value of a is

I=∫−π/2π/2(x2cosx1+πx+1+sin2x1+esinx2023)dx I=∫−π/2π/2(x2cosx1+π−x+1+sin2x1+esin(−x)2023)dx On Adding, we get 2I=∫−π/2π/2(x2cosx+1+sin2x)dxOn solving I=π24+3π4−2 a=3

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If the value of the integral $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} \cos x}{1 + π^{x}} + \frac{1 + \sin^{2} x}{1 + e^{\sin x^{2023}}}) d x = \frac{π}{4} (π + a) - 2,$ then the value of $a$ is

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$- \frac{3}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

$I = \int_{- π / 2}^{π / 2} (\frac{x^{2} \cos x}{1 + π^{x}} + \frac{1 + \sin^{2} x}{1 + e^{\sin x^{2023}}}) d x I = \int_{- π / 2}^{π / 2} (\frac{x^{2} \cos x}{1 + π^{- x}} + \frac{1 + \sin^{2} x}{1 + e^{\sin {(- x)}^{2023}}}) d x$

On Adding, we get
$2 I = \int_{- π / 2}^{π / 2} (x^{2} \cos x + 1 + \sin^{2} x) d x$
On solving
$I = \frac{π^{2}}{4} + \frac{3 π}{4} - 2 a = 3$