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Author: Tinku Tara

Question-71759

Question Number 71759 by naka3546 last updated on 19/Oct/19 Commented by Prithwish sen last updated on 19/Oct/19 $$\boldsymbol{\mathrm{x}}=\:\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{2}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{5}}………\frac{\mathrm{998}}{\mathrm{1000}}.\frac{\mathrm{999}}{\mathrm{1001}}\:=\:\frac{\mathrm{2}×\mathrm{1001001}}{\mathrm{3×1000}×\mathrm{1001}}\approx\mathrm{0}.\mathrm{67} \\ $$$$\therefore\:\mathrm{100x}\:\approx\:\mathrm{67} \\ $$$$\boldsymbol{\mathrm{By}}\:\boldsymbol{\mathrm{considering}} \\ $$$$\frac{\boldsymbol{\mathrm{n}}^{\mathrm{3}} −\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{3}}…

tanx-dx-

Question Number 6220 by sanusihammed last updated on 18/Jun/16 $$\int\sqrt{{tanx}}\:\:{dx}\: \\ $$ Answered by malwaan last updated on 19/Jun/16 $$\int\sqrt{{tanx}}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{1}+\sqrt{\mathrm{2}{tanx}}\right)\right. \\ $$$$−\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{1}−\sqrt{\mathrm{2}{tanx}}\right) \\…

5-2cos-2x-4-2sin-2x-dx-

Question Number 137285 by bemath last updated on 31/Mar/21 $$\int\:\frac{\mathrm{5}+\mathrm{2cos}\:\mathrm{2x}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:\mathrm{dx}\: \\ $$ Answered by EDWIN88 last updated on 31/Mar/21 $$\mathrm{E}=\int\:\frac{\mathrm{2cos}\:\mathrm{2x}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:\mathrm{dx}\:+\:\int\:\frac{\mathrm{5}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:\mathrm{dx} \\ $$$$\mathrm{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{d}\left(\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}\right)}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:+\:\int\:\frac{\mathrm{5}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:\mathrm{dx} \\ $$$$\mathrm{E}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}\:\mid\:+\:\int\:\frac{\mathrm{5}}{\mathrm{4}+\mathrm{2sin}\:\mathrm{2x}}\:\mathrm{dx}\: \\…

tanx-dx-

Question Number 6214 by sanusihammed last updated on 18/Jun/16 $$\int\sqrt{{tanx}}\:\:{dx}\: \\ $$ Commented by nburiburu last updated on 24/Jun/16 $${by}\:\:{substitution} \\ $$$${t}=\sqrt{{tan}\:{x}}\Rightarrow{t}^{\mathrm{2}} =\:{tan}\:{x} \\ $$$$\mathrm{2}{t}\:{dt}\:=\:{sec}^{\mathrm{2}}…