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Question-200894

Question Number 200894 by sonukgindia last updated on 26/Nov/23 Answered by Frix last updated on 26/Nov/23 $$=\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{4sin}^{\mathrm{2}} \:{x}}\:\overset{{t}=\mathrm{tan}\:{x}} {=} \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}}…

Question-200934

Question Number 200934 by sonukgindia last updated on 26/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:\pi{x}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} {dx}\:\overset{{t}=\left(\mathrm{cot}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } {=} \\ $$$$=−\frac{\mathrm{5}}{\pi}\underset{\infty} {\overset{\mathrm{0}}…

Question-200925

Question Number 200925 by akolade last updated on 26/Nov/23 Answered by mr W last updated on 26/Nov/23 $$\boldsymbol{{p}}=\overset{\rightarrow} {{AB}}=\left(\mathrm{6},\:−\mathrm{14}\right) \\ $$$${C}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{2}.\mathrm{5},\:\mathrm{1}.\mathrm{5}\right) \\ $$$${D}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{2}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{4},\:−\mathrm{2}\right) \\ $$$${E}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{5}.\mathrm{5},\:−\mathrm{5}.\mathrm{5}\right)…

Question-200903

Question Number 200903 by MrGHK last updated on 26/Nov/23 Answered by witcher3 last updated on 26/Nov/23 $$\mathrm{Hint} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\Psi\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{n}\right)\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{2n}}\right)+\Psi\left(\mathrm{2n}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\mathrm{2n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}ln}\left(\underset{\mathrm{1}} {\overset{\mathrm{x}}…

Question-200861

Question Number 200861 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $${J}=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} } \frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{5}}{\mathrm{1}+{t}}{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty}…

Question-200862

Question Number 200862 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $$\Phi \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx}=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} } \frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}}…

Question-200863

Question Number 200863 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $${x}^{\mathrm{2}\pi} ={t}\:\Rightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}} \:{and} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }{dx}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}}…