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Category: Integration

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Question Number 200254 by mnjuly1970 last updated on 16/Nov/23 $$ \\ $$$$\:\:\:\:\:\:{calculate}\:… \\ $$$$\:\:\Omega\:=\:\int_{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{tan}\left({x}\right)\right){dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }\:{dx}} \mathrm{ln}\left(\mathrm{sin}\left({x}\right)\right){dx}=? \\ $$ Answered…

Question-200250

Question Number 200250 by Calculusboy last updated on 16/Nov/23 Answered by Mathspace last updated on 16/Nov/23 $$\int_{\mathrm{0}} ^{{x}} {sin}\left({x}+{t}\right){dt} \\ $$$$=\left[{cos}\left({x}+{t}\right)\right]_{{t}=\mathrm{0}} ^{{t}={x}} ={cos}\left(\mathrm{2}{x}\right)−{cosx} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left(\int_{\mathrm{0}}…

Question-200299

Question Number 200299 by Calculusboy last updated on 16/Nov/23 Answered by witcher3 last updated on 17/Nov/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\Omega=\int_{\infty} ^{\mathrm{0}} −\frac{\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} }.\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}}…

Question-200257

Question Number 200257 by Calculusboy last updated on 16/Nov/23 Answered by witcher3 last updated on 16/Nov/23 $$\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}=\mathrm{y}\Rightarrow\mathrm{dy}=\mathrm{2}\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}} \\ $$$$=\int\mathrm{y}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}}…

Question-200159

Question Number 200159 by Calculusboy last updated on 15/Nov/23 Commented by 0670322918 last updated on 15/Nov/23 $$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{dx}= \\…