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Author: Tinku Tara

Question-212024

Question Number 212024 by Spillover last updated on 27/Sep/24 Answered by Frix last updated on 27/Sep/24 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} {dx}\:\overset{\left[{t}=\left(\mathrm{tan}\:{x}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]} {=}\:\mathrm{3}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{6}} +\mathrm{1}}…

Question-212023

Question Number 212023 by Spillover last updated on 27/Sep/24 Answered by Frix last updated on 27/Sep/24 $$\int\:\frac{\left({x}+\mathrm{1}\right)\mathrm{tan}\:{x}}{\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}+\mathrm{1}−\frac{\mathrm{1}}{\underset{\left[{t}=\mathrm{tan}\:{x}\right]} {\underbrace{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}}}−\frac{{x}}{\underset{\left[\mathrm{by}\:\mathrm{parts}\right]} {\underbrace{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}}}\right){dx}= \\ $$$$… \\…

Question-211989

Question Number 211989 by Spillover last updated on 26/Sep/24 Answered by som(math1967) last updated on 26/Sep/24 $$\:{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta+{cos}\beta{cos}\gamma \\ $$$$+{sin}\beta{sin}\gamma+{cos}\gamma{cos}\alpha+{sin}\gamma{sin}\alpha \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({cos}\alpha{cos}\beta+{cos}\beta{cos}\gamma+{cos}\gamma{cos}\alpha\right) \\ $$$$+\mathrm{2}\left({sin}\alpha{sin}\beta+{sin}\beta{sin}\gamma+{sin}\alpha{sin}\gamma\right)…