∫−22sin−1sinx+cos−1cosx1+x21+x25 dx=(where [. ] represents the greatest integer function)

I=∫−22 sin−1(sinx)+cos−1(cosx)1+x21+x25dxsin−1(sinx) is an odd function Also −2<x<2⇒0≤x2≤4⇒x25=0 ∴I=∫−22 cos−1(cosx)1+x2dx=2∫02 cos−1(cosx)1+x2dx ∵∫−aa f(x)dx=2∫0a f(x) if f(x) iseven =∫02 2x1+x2dx since cos-1cosx=x if x∈0,π =log1+x202 =log(1+4)−log(1) =loge5

∫−22sin−1sinx+cos−1cosx1+x21+x25 dx=(where [. ] represents the greatest integer function)

I=∫−22 sin−1(sinx)+cos−1(cosx)1+x21+x25dxsin−1(sinx) is an odd function Also −2<x<2⇒0≤x2≤4⇒x25=0 ∴I=∫−22 cos−1(cosx)1+x2dx=2∫02 cos−1(cosx)1+x2dx ∵∫−aa f(x)dx=2∫0a f(x) if f(x) iseven =∫02 2x1+x2dx since cos-1cosx=x if x∈0,π =log1+x202 =log(1+4)−log(1) =loge5

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

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$\log_{e} 2$

$\log_{e} 3$

$\log_{e} 5$

$\log_{e} 4$

answer is C.

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Detailed Solution

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