2} + {x^4}} \RIGHT)} \right]dx\) Let us take, x2 = tDifferentiating ‘x’ with respect to ‘t’:\( \Rightarrow 2x\FRAC{{dx}}{{DT}} = 1\) ⇒ 2x dx = dt\( \Rightarrow xdx = \frac{1}{2}{\rm{\;}}dt\) Then,\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\rm{\;co}}{{\rm{t}}^{ - 1}}{\rm{\;}}\left( {1 - t + {t^2}} \right)dt\) We need to convert cot x to tan x for easy calculation.\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\frac{1}{{\left( {1 - t + {t^2}} \right)}}\) By using the formula,∵ \({\tan ^{ - 1}}x - {\tan ^1}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)In this problem, (x = t); (y = t – 1)\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {\frac{{t - \left( {t - 1} \right)}}{{1 + t\left( {t - 1} \right)}}} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 TA{n^{ - 1}}{\rm{\;}}t - ta{n^{ - 1}}\left( {t - 1} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {t - 1} \right)dt} \right]{\rm{\;}}\) By using the formula,\(\mathop \smallint \limits_0^a f\left( x \right)dx = \mathop \smallint \limits_0^a f\left( {a - x} \right){\rm{\;}}dx\) Then, it becomes\( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {1 - t - 1} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( { - t} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt + \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}tdt} \right]\) ∵ tan-1 (-θ) = -tan-1 (θ)\( \Rightarrow I = \frac{1}{2}\left[ {2\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt} \right]\) \( \Rightarrow I = \mathop \smallint \limits_0^1 {\tan ^{ - 1}}{\rm{\;}}t{\rm{\;}}dt\) \( \Rightarrow I = \left[ {t.{{\tan }^{ - 1}}t - \frac{1}{2}{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {t.ta{n^{ - 1}}t} \right]_0^1 - \frac{1}{2}\left[ {{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {\left( {1.{{\tan }^{ - 1}}1} \right) - \left( {0.{{\tan }^{ - 1}}0} \right)} \right] - \frac{1}{2}\left[ {\left( {{{\log }_e}\left( {1 + 1} \right)} \right) - \left( {{{\log }_e}\left( {1 + 0} \right)} \right)} \right]\) \( \Rightarrow I = {\tan ^{ - 1}}1 - \frac{1}{2}{\log _e}\left( 2 \right)\) ∵ \({\tan ^{ - 1}}1 = 45^\circ = \frac{\pi }{4}\)\(\therefore I = \frac{\pi }{4} - \frac{1}{2}{\log _e}\left( 2 \right)\)

"> 2} + {x^4}} \RIGHT)} \right]dx\) Let us take, x2 = tDifferentiating ‘x’ with respect to ‘t’:\( \Rightarrow 2x\FRAC{{dx}}{{DT}} = 1\) ⇒ 2x dx = dt\( \Rightarrow xdx = \frac{1}{2}{\rm{\;}}dt\) Then,\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\rm{\;co}}{{\rm{t}}^{ - 1}}{\rm{\;}}\left( {1 - t + {t^2}} \right)dt\) We need to convert cot x to tan x for easy calculation.\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\frac{1}{{\left( {1 - t + {t^2}} \right)}}\) By using the formula,∵ \({\tan ^{ - 1}}x - {\tan ^1}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)In this problem, (x = t); (y = t – 1)\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {\frac{{t - \left( {t - 1} \right)}}{{1 + t\left( {t - 1} \right)}}} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 TA{n^{ - 1}}{\rm{\;}}t - ta{n^{ - 1}}\left( {t - 1} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {t - 1} \right)dt} \right]{\rm{\;}}\) By using the formula,\(\mathop \smallint \limits_0^a f\left( x \right)dx = \mathop \smallint \limits_0^a f\left( {a - x} \right){\rm{\;}}dx\) Then, it becomes\( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {1 - t - 1} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( { - t} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt + \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}tdt} \right]\) ∵ tan-1 (-θ) = -tan-1 (θ)\( \Rightarrow I = \frac{1}{2}\left[ {2\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt} \right]\) \( \Rightarrow I = \mathop \smallint \limits_0^1 {\tan ^{ - 1}}{\rm{\;}}t{\rm{\;}}dt\) \( \Rightarrow I = \left[ {t.{{\tan }^{ - 1}}t - \frac{1}{2}{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {t.ta{n^{ - 1}}t} \right]_0^1 - \frac{1}{2}\left[ {{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {\left( {1.{{\tan }^{ - 1}}1} \right) - \left( {0.{{\tan }^{ - 1}}0} \right)} \right] - \frac{1}{2}\left[ {\left( {{{\log }_e}\left( {1 + 1} \right)} \right) - \left( {{{\log }_e}\left( {1 + 0} \right)} \right)} \right]\) \( \Rightarrow I = {\tan ^{ - 1}}1 - \frac{1}{2}{\log _e}\left( 2 \right)\) ∵ \({\tan ^{ - 1}}1 = 45^\circ = \frac{\pi }{4}\)\(\therefore I = \frac{\pi }{4} - \frac{1}{2}{\log _e}\left( 2 \right)\)

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The value of the integral \(\mathop \smallint \nolimits_0^1 x\;co{t^{ - 1}}\left( {1 - {x^2} + {x^4}} \right)dx\) is:

General Knowledge General Awareness in General Knowledge 8 months ago

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Given,\(I = \mathop \smallint \limits_0^1 \left[ {x{\rm{\;}}co{t^{ - 1}}{\rm{\;}}\left( {1 - {x^2} + {x^4}} \RIGHT)} \right]dx\) Let us take, x2 = tDifferentiating ‘x’ with respect to ‘t’:\( \Rightarrow 2x\FRAC{{dx}}{{DT}} = 1\) ⇒ 2x dx = dt\( \Rightarrow xdx = \frac{1}{2}{\rm{\;}}dt\) Then,\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\rm{\;co}}{{\rm{t}}^{ - 1}}{\rm{\;}}\left( {1 - t + {t^2}} \right)dt\) We need to convert cot x to tan x for easy calculation.\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\frac{1}{{\left( {1 - t + {t^2}} \right)}}\) By using the formula,∵ \({\tan ^{ - 1}}x - {\tan ^1}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)In this problem, (x = t); (y = t – 1)\( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {\frac{{t - \left( {t - 1} \right)}}{{1 + t\left( {t - 1} \right)}}} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\mathop \smallint \limits_0^1 TA{n^{ - 1}}{\rm{\;}}t - ta{n^{ - 1}}\left( {t - 1} \right)dt{\rm{\;}}\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {t - 1} \right)dt} \right]{\rm{\;}}\) By using the formula,\(\mathop \smallint \limits_0^a f\left( x \right)dx = \mathop \smallint \limits_0^a f\left( {a - x} \right){\rm{\;}}dx\) Then, it becomes\( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( {1 - t - 1} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt - \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}\left( { - t} \right)dt} \right]\) \( \Rightarrow I = \frac{1}{2}\left[ {\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt + \mathop \smallint \limits_0^1 {{\tan }^{ - 1}}tdt} \right]\) ∵ tan-1 (-θ) = -tan-1 (θ)\( \Rightarrow I = \frac{1}{2}\left[ {2\mathop \smallint \limits_0^1 {{\tan }^{ - 1}}{\rm{\;}}t{\rm{\;}}dt} \right]\) \( \Rightarrow I = \mathop \smallint \limits_0^1 {\tan ^{ - 1}}{\rm{\;}}t{\rm{\;}}dt\) \( \Rightarrow I = \left[ {t.{{\tan }^{ - 1}}t - \frac{1}{2}{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {t.ta{n^{ - 1}}t} \right]_0^1 - \frac{1}{2}\left[ {{{\log }_e}\left( {1 + {t^2}} \right)} \right]_0^1\) \( \Rightarrow I = \left[ {\left( {1.{{\tan }^{ - 1}}1} \right) - \left( {0.{{\tan }^{ - 1}}0} \right)} \right] - \frac{1}{2}\left[ {\left( {{{\log }_e}\left( {1 + 1} \right)} \right) - \left( {{{\log }_e}\left( {1 + 0} \right)} \right)} \right]\) \( \Rightarrow I = {\tan ^{ - 1}}1 - \frac{1}{2}{\log _e}\left( 2 \right)\) ∵ \({\tan ^{ - 1}}1 = 45^\circ = \frac{\pi }{4}\)\(\therefore I = \frac{\pi }{4} - \frac{1}{2}{\log _e}\left( 2 \right)\)

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Posted on 23 Nov 2024, this text provides information on General Knowledge related to General Awareness in General Knowledge. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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