If the value of the integral - / 2 / 2 ( x 2 cos x 1 + x + 1 + sin 2 x 1 + e sin x 2023 ) d x = 4 ( + a ) - 2 , then the value of a is [2024]

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Solution

Q.
If the value of the integral $\int_{- π / 2}^{π / 2} (\frac{x^{2} \cos x}{1 + π^{x}} + \frac{1 + \sin^{2} x}{1 + {e^{\sin x}}^{2023}}) d x = \frac{π}{4} (π + a) - 2,$ then the value of $a$ is [2024]

1 $2$

2 $- \frac{3}{2}$

3 $3$

4 $\frac{3}{2}$

Ans.
(3)

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

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

Adding (i) and (ii), we get

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

$\Rightarrow I = \int_{0}^{π / 2} (x^{2} \cos x + 1 + \sin^{2} x) d x$

$= \int_{0}^{π / 2} x^{2} \cos x d x + \int_{0}^{π / 2} (1 + \sin^{2} x) d x$ ...(iii)

Let $I_{1} = \int_{0}^{π / 2} x^{2} \cos x d x = {[x^{2} (\sin x) - \int 2 x \sin x d x]}_{0}^{π / 2}$

$= {[x^{2} \sin x + 2 x \cos x - 2 \sin x]}_{0}^{π / 2}$

$I_{1} = {(\frac{π}{2})}^{2} \sin \frac{π}{2} + 2 \frac{π}{2} \cos \frac{π}{2} - 2 \sin \frac{π}{2} - 0 = \frac{π^{2}}{4} - 2$

And $I_{2} = \int_{0}^{π / 2} (1 + \sin^{2} x) d x = \int_{0}^{π / 2} [1 + (\frac{1 - \cos 2 x}{2})] d x$

$= {[\frac{3}{2} x - \frac{1}{4} \sin 2 x]}_{0}^{π / 2} = \frac{3}{2} \times \frac{π}{2} - \frac{1}{4} \sin 2 \times \frac{π}{2} = \frac{3 π}{4}$

From (iii), $I = \frac{π^{2}}{4} - 2 + \frac{3 π}{4} = \frac{π}{4} (π + 3) - 2$

Comparing with $\frac{π}{4} (π + a) - 2, we get a = 3$

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