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

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(A) माना $\alpha=\int_{\frac{1}{2}}^2 \frac{	an ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करत

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) माना $\alpha=\int_{\frac{1}{2}}^2 \frac{	an ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करते हुए,या $x = \frac{1}{t}$ प्रतिस्थापन लेने पर,$dx = -\frac{1}{t^2} dt$ प्राप्त होता है।जब $x = \frac{1}{2}, t = 2$ और जब $x = 2, t = \frac{1}{2}$.$\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} x}{2x^2 - 3x + 2} dx \quad ...(ii)$$(i)$ और $(ii)$ को जोड़ने पर:$2\alpha = \int_{\frac{1}{2}}^2 \frac{	an ^{-1} x + \cot ^{-1} x}{2x^2 - 3x + 2} dx = \int_{\frac{1}{2}}^2 \frac{\pi/2}{2x^2 - 3x + 2} dx$$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} (3\sqrt{7})$अतः,$\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$.

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