Question Number 52893 by dwdkswd last updated on 14/Jan/19 $$\int\mathrm{sin}\:{x}×\mathrm{cos}\:{x}\:{dx} \\ $$ Answered by MJS last updated on 14/Jan/19 $$\int\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{2}{x}\:{dx}=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}{x}\:+{C}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \:{x}\:+{C}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \:{x}\:+{C} \\…
Question Number 118419 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{4cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\: \\ $$ Commented by bramlexs22 last updated on 17/Oct/20 $${yes}…{thank}\:{you}\:{all}\:{master}\: \\ $$ Answered by peter…
Question Number 118395 by TANMAY PANACEA last updated on 17/Oct/20 $$\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$ Commented by Dwaipayan Shikari last updated on 17/Oct/20 $$\frac{{b}−{a}}{\mathrm{2}} \\…
Question Number 118347 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\int\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt[{\mathrm{3}\:}]{{x}}}\: \\ $$ Answered by benjo_mathlover last updated on 17/Oct/20 $${setting}\:{x}\:=\:\:\:{r}^{\mathrm{6}} \\ $$$$\int\:\frac{\mathrm{6}{r}^{\mathrm{5}} }{{r}^{\mathrm{3}} +{r}^{\mathrm{2}} }\:{dr}\:=\:\:\int\frac{\mathrm{6}{r}^{\mathrm{3}}…
Question Number 183879 by Michaelfaraday last updated on 31/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 118338 by benjo_mathlover last updated on 17/Oct/20 $$\:\:\:\:{solve}\:\int\:\frac{{dx}}{\mathrm{3}−\mathrm{5sin}\:{x}}\: \\ $$ Answered by Dwaipayan Shikari last updated on 17/Oct/20 $$\int\frac{{dx}}{\mathrm{3}−\mathrm{5}{sinx}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{3}−\frac{\mathrm{10}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}}…
Question Number 118340 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\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}…