SATHEE: Indefinite Integration 1 Question 17

Indefinite Integration 1 Question 17

18. The value of $\int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$ is equal to

(a) $\frac{π^{2}}{4} - 2$

(b) $\frac{Π^{2}}{4} + 2$

(d) $π^{2} + e^{π / 2}$

(2016 Adv.)

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Solution:

Let $I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$

$\begin{aligned} ∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ \Rightarrow I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos (- x)}{1 + e^{- x}} d x \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I = \int_{- π / 2}^{π / 2} x^{2} \cos x \frac{1}{1 + e^{x}} + \frac{1}{1 + e^{- x}} d x \\ = \int_{- π / 2}^{π / 2} x^{2} \cos x \cdot (1) d x \\ ∵ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x, when f (- x) = f (x) \\ \Rightarrow 2 I = 2 \int_{0}^{π / 2} x^{2} \cos x d x \end{aligned}$

Using integration by parts, we get

$\begin{aligned} 2 I & = 2 {[x^{2} (\sin x) - (2 x) (- \cos x) + (2) (- \sin x)]}_{0}^{π / 2} \\ \Rightarrow 2 I & = 2 \frac{π^{2}}{4} - 2 \\ ∴ I & = \frac{π^{2}}{4} - 2 \end{aligned}$

Indefinite Integration 1 Question 17

18. The value of $\int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$ is equal to

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Indefinite Integration 1 Question 17

18. The value of ∫−π/2π/2x2cos⁡x1+exdx is equal to

Solution:

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18. The value of $\int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$ is equal to