SATHEE: Indefinite Integration 1 Question 21

Indefinite Integration 1 Question 21

22. The value of the integral $\int_{- π / 2}^{π / 2} x^{2} + \log \frac{π - x}{π + x} \cos x d x$ is

(a) 0

(b) $\frac{π^{2}}{2} - 4$

(d) $\frac{π^{2}}{2}$

(2012)

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Solution:

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

As, $\int_{- a}^{a} f (x) d x = 0$ , when $f (- x) = - f (x)$

$\begin{aligned} ∴ I & = \int_{- π / 2}^{π / 2} x^{2} \cos x d x + 0 = 2 \int_{0}^{π / 2} (x^{2} \cos x) d x \\ = 2 {(x^{2} \sin x)}_{0}^{π / 2} - \int_{0}^{π / 2} 2 x \cdot \sin x d x \\ = 2 \frac{π^{2}}{4} - 2 (- x \cdot \cos x)_{0}^{π / 2} - \int_{0}^{π / 2} 1 \cdot (- \cos x) d x \\ = 2 \frac{π^{2}}{4} - 2 (\sin x)_{0}^{π / 2} = 2 \frac{π^{2}}{4} - 2 = \frac{π^{2}}{2} - 4 \end{aligned}$

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Indefinite Integration 1 Question 21

22. The value of the integral ∫−π/2π/2x2+log⁡π−xπ+xcos⁡xdx is

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22. The value of the integral $\int_{- π / 2}^{π / 2} x^{2} + \log \frac{π - x}{π + x} \cos x d x$ is