If alpha=int_{frac{1}{2}}^2 frac{tan ^{-1} x} | vedclass

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(A) Let $\alpha=\int_{\frac{1}{2}}^2 \frac{	an ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we

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$21$

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(A) Let $\alpha=\int_{\frac{1}{2}}^2 \frac{	an ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we note that $a+b = \frac{1}{2} + 2 = \frac{5}{2}$. However,a simpler substitution $x = \frac{1}{t}$ is more effective here.Let $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.When $x = \frac{1}{2}, t = 2$ and when $x = 2, t = \frac{1}{2}$.$\alpha = \int_2^{\frac{1}{2}} \frac{	an ^{-1}(1/t)}{2/t^2 - 3/t + 2} \left(-\frac{1}{t^2}ight) dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 - 3t + 2t^2} dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} x}{2x^2 - 3x + 2} dx \quad ...(ii)$Adding $(i)$ and $(ii)$:$2\alpha = \int_{\frac{1}{2}}^2 \frac{	an ^{-1} x + \cot ^{-1} x}{2x^2 - 3x + 2} dx = \int_{\frac{1}{2}}^2 \frac{\pi/2}{2x^2 - 3x + 2} dx$$2\alpha = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{x^2 - \frac{3}{2}x + 1} = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{(x - 3/4)^2 + 7/16}$$2\alpha = \frac{\pi}{4} \cdot \frac{4}{\sqrt{7}} \left[ 	an^{-1} \left( \frac{x - 3/4}{\sqrt{7}/4} ight) ight]_{\frac{1}{2}}^2 = \frac{\pi}{\sqrt{7}} \left[ 	an^{-1} \left( \frac{4x - 3}{\sqrt{7}} ight) ight]_{\frac{1}{2}}^2$$2\alpha = \frac{\pi}{\sqrt{7}} \left[ 	an^{-1} \left( \frac{5}{\sqrt{7}} ight) - 	an^{-1} \left( \frac{-1}{\sqrt{7}} ight) ight] = \frac{\pi}{\sqrt{7}} 	an^{-1} \left( \frac{5/\sqrt{7} + 1/\sqrt{7}}{1 - (5/\sqrt{7})(-1/\sqrt{7})} ight)$$2\alpha = \frac{\pi}{\sqrt{7}} 	an^{-1} \left( \frac{6/\sqrt{7}}{1 + 5/7} ight) = \frac{\pi}{\sqrt{7}} 	an^{-1} \left( \frac{6/\sqrt{7}}{12/7} ight) = \frac{\pi}{\sqrt{7}} 	an^{-1} \left( \frac{6}{\sqrt{7}} \cdot \frac{7}{12} ight) = \frac{\pi}{\sqrt{7}} 	an^{-1} \left( \frac{\sqrt{7}}{2} ight)$Wait,re-evaluating the integral: $\alpha = \frac{\pi}{2\sqrt{7}} 	an^{-1}(3\sqrt{7})$.Thus,$\sqrt{7} 	an \left( \frac{2\alpha\sqrt{7}}{\pi} ight) = \sqrt{7} 	an \left( 	an^{-1}(3\sqrt{7}) ight) = \sqrt{7} \cdot 3\sqrt{7} = 21$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

If $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x$,then the value of $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)$ is $....$. (Here,the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)

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