Question Number 118340 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\underset{\pi/\mathrm{3}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$ Commented by Dwaipayan Shikari last updated on 17/Oct/20 $$\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}−{cosx}+{sinx}}=\mathrm{2}\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}…
Question Number 183865 by Michaelfaraday last updated on 31/Dec/22 Answered by MJS_new last updated on 31/Dec/22 $$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mid\mathrm{1}−\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mid{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{3}} {\int}}\left(\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\:−\mathrm{1}\right){dx}+\underset{\pi/\mathrm{3}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\right){dx}=…
Question Number 118318 by mnjuly1970 last updated on 16/Oct/20 Answered by Bird last updated on 17/Oct/20 $${A}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{4}} {x}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${A}\:=\left[−\frac{{sin}^{\mathrm{4}} {x}}{{x}}\right]_{\mathrm{0}} ^{\infty}…
Question Number 183844 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 $$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{x}−\mathrm{1}}+\sqrt{{x}−\mathrm{3}}}= \\ $$$$=\int\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}−\sqrt{{x}−\mathrm{3}}}{\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}+\sqrt{{x}−\mathrm{3}}\right)\left(\sqrt{\mathrm{2}{x}−\mathrm{1}−}\sqrt{{x}−\mathrm{3}}\right)}{dx}= \\ $$$$=\int\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{{x}+\mathrm{2}}{dx}−\int\frac{\sqrt{{x}−\mathrm{3}}}{{x}+\mathrm{2}}{dx}= \\ $$$$\mathrm{with}\:{u}=\sqrt{\mathrm{2}{x}−\mathrm{1}}\:\mathrm{and}\:{v}=\sqrt{{x}−\mathrm{3}}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{get} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{5}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\:\sqrt{\mathrm{5}}}\:−\mathrm{2}\sqrt{{x}−\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}\mathrm{arctan}\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\:+{C}…
Question Number 183845 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}{{t}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\:\sqrt{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{2}{t}+\sqrt{\mathrm{2}}}=\mathrm{2arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:=…
Question Number 118307 by bramlexs22 last updated on 16/Oct/20 $$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$ Commented by Lordose last updated on 16/Oct/20 $$\mathrm{see}\:\mathrm{qst}\:\mathrm{118030}…
Question Number 118292 by Lordose last updated on 16/Oct/20 $$\boldsymbol{\mathrm{P}}\mathrm{rove}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{lnx}}\:=\:\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{n}}+\mathrm{1}\mid \\ $$ Answered by TANMAY PANACEA last updated on 16/Oct/20…
Question Number 118278 by Lordose last updated on 16/Oct/20 $$\mathrm{Evaluate} \\ $$$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{3}}} \mathrm{tan}^{\mathrm{2}} \mathrm{xsec}\left(\frac{\mathrm{x}}{\mathrm{3}}\right)\mathrm{dx} \\ $$$$\bigstar \\ $$ Answered by MJS_new last updated on…
Question Number 118270 by bramlexs22 last updated on 16/Oct/20 $$\:\:\:\int{x}\:{a}^{{x}} \:\left(\mathrm{1}−{a}\right)^{\mathrm{1}−{x}} \:{dx}? \\ $$ Commented by Dwaipayan Shikari last updated on 16/Oct/20 $$\left(\mathrm{1}−{a}\right)\int{x}\left(\frac{{a}}{\left(\mathrm{1}−{a}\right)}\right)^{{x}} \\ $$$$\frac{\mathrm{1}−{a}}{{log}\left(\frac{{a}}{\mathrm{1}−{a}}\right)}{x}\left(\frac{{a}}{\mathrm{1}−{a}}\right)^{{x}}…
Question Number 183806 by cortano1 last updated on 30/Dec/22 $$\:\:\int\:\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}}{{x}}\:{dx}\:=? \\ $$ Answered by Ar Brandon last updated on 30/Dec/22 $${I}=\int\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}}{{x}}{dx}\:,\:{x}=\mathrm{5sh}\vartheta \\ $$$$\:\:=\int\frac{\sqrt{\mathrm{5sh}\vartheta+\mathrm{5ch}\vartheta}}{\mathrm{5sh}\vartheta}\left(\mathrm{5ch}\vartheta{d}\vartheta\right)=\sqrt{\mathrm{5}}\int\left(\frac{{e}^{\mathrm{2}\vartheta}…