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Author: Tinku Tara

show-that-0-pi-2-ln-sec-x-ln-csc-x-dx-pi-2-ln-2-2-2-pi-4-48-

Question Number 89593 by M±th+et£s last updated on 19/Apr/20 $${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sec}\left({x}\right)\right)\:{ln}\left({csc}\left({x}\right)\right)\:{dx}=\frac{\pi^{\mathrm{2}} \:{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{4}} }{\mathrm{48}} \\ $$ Commented by maths mind last updated…

Compare-the-bond-strength-of-S-O-bond-in-SO-3-2-and-SO-4-2-ion-

Question Number 24054 by Tinkutara last updated on 12/Nov/17 $$\mathrm{Compare}\:\mathrm{the}\:\mathrm{bond}\:\mathrm{strength}\:\mathrm{of}\:\mathrm{S}\:−\:\mathrm{O} \\ $$$$\mathrm{bond}\:\mathrm{in}\:\mathrm{SO}_{\mathrm{3}} ^{−\mathrm{2}} \:\mathrm{and}\:\mathrm{SO}_{\mathrm{4}} ^{−\mathrm{2}} \:\mathrm{ion}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

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

Question Number 89584 by jagoll last updated on 18/Apr/20 $$\int\:\frac{\mathrm{sin}\:^{\mathrm{4}} \left(\mathrm{x}\right)\:\mathrm{dx}}{\mathrm{4}+\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$ Commented by mathmax by abdo last updated on 18/Apr/20 $${I}\:=\int\:\:\frac{{sin}^{\mathrm{4}} {xdx}}{\mathrm{4}+{cos}^{\mathrm{2}}…

Question-155122

Question Number 155122 by saly last updated on 25/Sep/21 Commented by benhamimed last updated on 25/Sep/21 $$/{x}−\mathrm{3}/+\mathrm{3}−{x}\neq\mathrm{0} \\ $$$$/{x}−\mathrm{3}/=\begin{cases}{{x}−\mathrm{3}\:\:\:;{x}\geqslant\mathrm{3}}\\{\mathrm{3}−{x}\:\:\:\:;{x}\leqslant\mathrm{3}}\end{cases} \\ $$$$/{x}−\mathrm{3}/+\mathrm{3}−{x}=\begin{cases}{\mathrm{0}\:\:\:;{x}\geqslant\mathrm{3}}\\{\mathrm{6}−\mathrm{2}{x}\:\:\:;{x}\leqslant\mathrm{0}}\end{cases} \\ $$$$\left.{D}=\right]−\infty;\mathrm{3}\left[\right. \\ $$…

Question-89580

Question Number 89580 by jagoll last updated on 18/Apr/20 Answered by $@ty@m123 last updated on 18/Apr/20 $$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left\{\mathrm{180}−\left({x}+\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)}\:…\left(\mathrm{1}\right) \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{BC}}{\mathrm{sin}\:\left\{\mathrm{180}−\left(\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{AD}}{\mathrm{sin}\:\mathrm{54}}\:…\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right),…