First slide

Evaluation of definite integrals

Question

$\int_{- 2}^{2} \frac{\sin^{- 1} (\sin x) + \cos^{- 1} (\cos x)}{(1 + x^{2}) (1 + [\frac{x^{2}}{5}])} d x = (w h e r e [.] r e p r e s e n t s t h e g r e a t e s t i n t e g e r f u n c t i o n)$

Moderate

A

$\log_{e} 2$
B

$\log_{e} 3$
C

$\log_{e} 5$
D

$\log_{e} 4$

Solution

$\begin{array}{l} I = \int_{- 2}^{2} \frac{\sin^{- 1} (\sin x) + \cos^{- 1} (\cos x)}{(1 + x^{2}) (1 + [\frac{x^{2}}{5}])} d x \\ \sin^{- 1} (\sin x) is an odd function \\ Also - 2 < x < 2 \Rightarrow 0 \leq x^{2} \leq 4 \Rightarrow [\frac{x^{2}}{5}] = 0 \\ ∴ I = \int_{- 2}^{2} \frac{\cos^{- 1} (\cos x)}{1 + x^{2}} d x = 2 \int_{0}^{2} \frac{\cos^{- 1} (\cos x)}{1 + x^{2}} d x (∵ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) if f (x) iseven) \\ = \int_{0}^{2} \frac{2 x}{1 + x^{2}} d x (\sin c e \cos^{- 1} (\cos x) = x i f x \in [0, π]) \\ = {[\log (1 + x^{2})]}_{0}^{2} \\ = \log (1 + 4) - \log (1) \\ = \log_{e} 5 \end{array}$

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