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Author: Tinku Tara

0-pi-2-x-2-csc-2-x-dx-4z-2-e-iz-e-iz-2-dz-z-2-e-2iz-e-2iz-1-2-dz-u-Substitute-e-2iz-1-du-2ie-2iz-dz-z-2-ln-2-u-1-4-1-8-i-ln-

Question Number 221843 by wewji12 last updated on 11/Jun/25 $$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:{x}^{\mathrm{2}} \mathrm{csc}^{\mathrm{2}} \left({x}\right)\mathrm{d}{x} \\ $$$$\int\:\:−\frac{\mathrm{4}{z}^{\mathrm{2}} }{\left({e}^{\boldsymbol{{i}}{z}} −{e}^{−\boldsymbol{{i}}{z}} \right)^{\mathrm{2}} }\:\mathrm{d}{z}=\int\:\:\frac{{z}^{\mathrm{2}} {e}^{\mathrm{2}\boldsymbol{{i}}{z}} }{\left({e}^{\mathrm{2}\boldsymbol{{i}}{z}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{d}{z} \\…

Question-221870

Question Number 221870 by fantastic last updated on 11/Jun/25 Answered by mehdee7396 last updated on 12/Jun/25 $${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×{OM}\:\:\:\:\&\:\:\:{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{CD}×{ON} \\ $$$${AB}={CD}\:\:\Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×\left({OM}+{ON}\right)=\frac{{AB}×{MN}}{\mathrm{2}}=\mathrm{16} \\ $$$$\Rightarrow{S}={AB}×{MN}=\mathrm{32}…

0-arctan-x-x-2-dx-

Question Number 221838 by Ghisom last updated on 11/Jun/25 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{arctan}\:{x}}{{x}}\right)^{\mathrm{2}} {dx}=? \\ $$ Answered by wewji12 last updated on 11/Jun/25 $$\mathrm{tan}^{−\mathrm{1}} \left({t}\right)=\rho\:\rightarrow\:\mathrm{d}{t}=\mathrm{sec}^{\mathrm{2}} \left(\rho\right)\mathrm{d}\rho…

f-x-lim-h-0-f-x-h-f-x-h-lim-h-0-c-c-h-lim-h-0-0-0-lim-h-0-x-h-2-x-2-h-lim-h-0-2xh-h-2-h-lim-h-0-2x-h-2x-lim-h-0-x-h-n-x-n-h-lim-h-0-k-0-n-n

Question Number 221832 by MrGaster last updated on 11/Jun/25 $${f}'\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{c}−{c}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}0}=\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{xh}+{h}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}{x}+{h}\right)=\mathrm{2}{x} \\ $$$$\underset{{h}\rightarrow\mathrm{0}}…

70-71-72-73-1-

Question Number 221829 by fantastic last updated on 11/Jun/25 $$\sqrt{\mathrm{70}.\mathrm{71}.\mathrm{72}.\mathrm{73}+\mathrm{1}} \\ $$ Answered by aleks041103 last updated on 11/Jun/25 $$\mathrm{70}.\mathrm{71}.\mathrm{72}.\mathrm{73}+\mathrm{1}= \\ $$$$=\mathrm{70}.\mathrm{73}.\left(\mathrm{70}+\mathrm{1}\right)\left(\mathrm{73}−\mathrm{1}\right)+\mathrm{1}= \\ $$$$=\mathrm{70}.\mathrm{73}.\left(\mathrm{70}.\mathrm{73}+\mathrm{73}−\mathrm{70}−\mathrm{1}\right)+\mathrm{1}= \\…