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Category: Differentiation

Evaluate-0-pi-4-ln-tan-x-sin-pi-e-2x-sin-pi-e-x-cos-pi-e-x-2-dx-

Question Number 143086 by mnjuly1970 last updated on 09/Jun/21 $$ \\ $$$$\:\:{Evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({tan}\left({x}\right)\right).{sin}^{\pi^{{e}} } \left(\mathrm{2}{x}\right)}{\left({sin}^{\pi^{{e}} } \left({x}\right)+{cos}^{\pi^{{e}} } \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$…

Question-142935

Question Number 142935 by mnjuly1970 last updated on 07/Jun/21 Answered by qaz last updated on 07/Jun/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}}…

Differentiate-ln-cosx-from-the-first-principle-

Question Number 11843 by tawa last updated on 02/Apr/17 $$\mathrm{Differentiate},\:\:\mathrm{ln}\left(\mathrm{cosx}\right)\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{principle}. \\ $$ Answered by ajfour last updated on 02/Apr/17 $$\frac{{d}}{{dx}}\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\mathrm{cos}\:\left({x}+{h}\right)−\mathrm{ln}\:\mathrm{cos}\:{x}}{{h}} \\ $$$$\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\:\frac{\mathrm{cos}\:\left({x}+{h}\right)}{\mathrm{cos}\:{x}}\:\right]}{{h}} \\…

Question-142870

Question Number 142870 by mnjuly1970 last updated on 06/Jun/21 Answered by qaz last updated on 06/Jun/21 $$\Theta=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{u}}}{\mathrm{u}}\mathrm{du} \\ $$$$=\mathrm{2}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{u}}}{\mathrm{u}}\mathrm{du}+\int_{\mathrm{0}}…