Question Number 151067 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\centerdot\mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{x}}−\mathrm{1}\right)\mathrm{dx}=\sqrt{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151064 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\pi} \mathrm{arctan}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\varphi\mathrm{cos}\:\mathrm{x}+\mathrm{2}\varphi^{\mathrm{2}} }\right)\mathrm{dx}=\pi\mathrm{arctan}\:\sqrt{\varphi}\:\:\:\:\:\:\:\:\:\:\left(\varphi=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151061 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{x}\centerdot\mathrm{cot}\:\mathrm{x}\centerdot\mathrm{ln}^{\mathrm{2}} \mathrm{cos}\:\mathrm{xdx}=\frac{\pi^{\mathrm{3}} }{\mathrm{24}}\mathrm{ln2}+\frac{\pi}{\mathrm{6}}\mathrm{ln}^{\mathrm{3}} \mathrm{2}−\frac{\mathrm{3}}{\mathrm{16}}\pi\zeta\left(\mathrm{3}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151066 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\mathrm{arctan}\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2}}\right)+\mathrm{arctan}\left(\frac{\mathrm{cos}\:\mathrm{3x}+\mathrm{15cos}\:\mathrm{x}}{\mathrm{8}}\right)\right)\mathrm{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151062 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\centerdot\sqrt[{\mathrm{4}}]{\mathrm{8x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{1}}}=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{4}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151063 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\left(\pi^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \mathrm{x}\right)\mathrm{x}}\mathrm{dx}=\gamma \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151060 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:::\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}\sqrt{\mathrm{x}}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}+\mathrm{ax}\right)}\mathrm{dx}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{a}+\sqrt{\mathrm{2a}}}{\:\sqrt{\mathrm{2}}\mathrm{a}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}\pi\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:,\left(\mathrm{a}>\mathrm{0}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151053 by Tawa11 last updated on 17/Aug/21 Answered by puissant last updated on 18/Aug/21 $${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({lnx}\right)−{ln}\left({x}\right)}{{ln}^{\mathrm{2}} {x}}{dx} \\ $$$${t}=−{lnx}\:\rightarrow\:{x}={e}^{−{t}} \rightarrow{dx}=−{e}^{−{t}} {dt} \\…
Question Number 151037 by qaz last updated on 17/Aug/21 $$\underset{\mathrm{r}=\mathrm{1},\mathrm{r}\neq\mathrm{s}} {\overset{\mathrm{n}} {\sum}}\:\:\underset{\mathrm{s}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{rs}}{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}\overset{?} {=}\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{3n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$ Answered by mindispower last updated on 17/Aug/21 $$=\underset{{r}=\mathrm{1}}…
Question Number 85488 by john santu last updated on 22/Mar/20 $$\int\:\frac{{dx}}{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$ Answered by MJS last updated on 22/Mar/20 $$\mathrm{trying}\:\mathrm{around}\:\mathrm{I}\:\mathrm{landed}\:\mathrm{on} \\…