Question Number 100956 by bobhans last updated on 29/Jun/20 Answered by john santu last updated on 29/Jun/20 $$\mathrm{set}\:\mathrm{ln}\left(\mathrm{x}\right)\:=\:\mathrm{z}\:\rightarrow\begin{cases}{\mathrm{z}=\mathrm{1}}\\{\mathrm{z}=\mathrm{e}}\end{cases} \\ $$$$\left.\mathrm{I}=\:\underset{\mathrm{1}} {\overset{\mathrm{e}} {\int}}\:\mathrm{ln}\left(\mathrm{z}\right).\:\mathrm{z}\:\mathrm{dz}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}^{\mathrm{2}} .\mathrm{ln}\left(\mathrm{z}\right)\right]_{\mathrm{1}} ^{\mathrm{e}} −\frac{\mathrm{1}}{\mathrm{2}}\overset{\mathrm{e}}…
Question Number 166491 by mathlove last updated on 21/Feb/22 Commented by cortano1 last updated on 21/Feb/22 $$\:\:=\:\frac{\mathrm{1}}{\mathrm{16}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{64}}\:\mathrm{sin}\:\mathrm{4x}\:+\frac{\mathrm{1}}{\mathrm{48}}\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{2x}\:+\:\mathrm{c} \\ $$ Commented by mathlove last updated…
Question Number 166488 by bobhans last updated on 21/Feb/22 $$\:\:\:\mathrm{B}=\int\:\sqrt{\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{1}}{\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$ Answered by greogoury55 last updated on 21/Feb/22 $$\:\:\:{B}=\int\:\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}}\:{dx} \\ $$$$\:\:\:{B}=\int\:\sqrt{\frac{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{2sin}\:^{\mathrm{2}} {x}}}\:{dx}…
Question Number 100948 by Dwaipayan Shikari last updated on 29/Jun/20 Commented by Dwaipayan Shikari last updated on 29/Jun/20 $$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{r}}{{n}}\right){sec}^{\mathrm{2}} \left(\frac{{r}}{{n}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}}…
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Question Number 35379 by lilesh7 last updated on 18/May/18 $$\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\:? \\ $$ Commented by prof Abdo imad last updated on 18/May/18…
Question Number 166449 by cortano1 last updated on 20/Feb/22 $$\:\:\:\mathrm{C}\:=\:\int\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$ Commented by cortano1 last updated on 20/Feb/22 $$\mathrm{oo}\:\mathrm{yes} \\ $$ Commented…
Question Number 166436 by mnjuly1970 last updated on 20/Feb/22 $$ \\ $$$$\:\:\:\:\:\:\mathrm{S}{how}\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\:{n}} \:{H}_{\:{n}} }{{n}^{\:\mathrm{2}} }\:\:=\:−\frac{\mathrm{5}}{\mathrm{8}}\:\zeta\:\left(\mathrm{3}\:\right)\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:−−−−−−−−− \\ $$…
Question Number 35325 by Cheyboy last updated on 17/May/18 $${Sketch}\:{the}\:{region}\:{enclosed}\:{by}\:{the} \\ $$$${curves}\:{of}\:{y}=\mathrm{1}/{x}\:{and}\:{y}=\mathrm{1}/{x}^{\mathrm{2}} \:{and} \\ $$$${find}\:{the}\:{area}\:{of}\:{the}\:{region}. \\ $$$${plzz}\:{help}\:{me} \\ $$ Answered by MJS last updated on…
Question Number 166376 by mnjuly1970 last updated on 19/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com