Question Number 94286 by mhmd last updated on 17/May/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94278 by M±th+et+s last updated on 17/May/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{\sqrt{{x}}} }{\:\sqrt{{x}}}{dx} \\ $$ Commented by PRITHWISH SEN 2 last updated on 17/May/20 $$\mathrm{e}^{\sqrt{\mathrm{x}}}…
Question Number 94269 by LPM last updated on 17/May/20 Commented by prakash jain last updated on 17/May/20 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}}…
Question Number 94243 by mhmd last updated on 17/May/20 $${By}\:{using}\:{the}\:{double}\:{integral}\:{find}\:{the}\:{area}\:{of}\:{the}\:{regiin}\: \\ $$$${bounded}\:{by}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} +\mathrm{2}{x}\:,\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}\:{and}\:{x}−{axis}\: \\ $$$$\left({sketch}\:{the}\:{region}\:{of}\:{integration}\right) \\ $$$${pleas}\:{sir}\:{help}\:{me}\: \\ $$ Commented by Ar Brandon last…
Question Number 28705 by students last updated on 29/Jan/18 $${solve}\:{the}\:{integrayion}\:\frac{{sin}\mathrm{2}{x}}{{sin}\mathrm{5}{xsin}\mathrm{3}{x}} \\ $$ Answered by mrW2 last updated on 30/Jan/18 $$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\left(\mathrm{5}{x}−\mathrm{3}{x}\right)=\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\int\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{sin}\:\mathrm{3}{x}}{dx}=\int\mathrm{cot}\:\mathrm{3}{x}\:{dx}−\int\mathrm{cot}\:\mathrm{5}{x}\:{dx} \\ $$$$=\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{3}{x}\right)}{\mathrm{3}}−\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{5}{x}\right)}{\mathrm{5}}+{C} \\…
Question Number 28706 by students last updated on 29/Jan/18 $${most}\:{important}\:{question}\:{gor}\:{boar}\:{or}\:{iit}\: \\ $$$${solve}\:{the}\:{integration}\:\:\frac{\mathrm{1}}{\mathrm{3}{sinx}+\mathrm{4}{cosx}} \\ $$ Commented by abdo imad last updated on 29/Jan/18 $${let}\:{put}\:{I}=\:\int\:\:\frac{{dx}}{\mathrm{3}{sinx}+\mathrm{4}{cosx}}\:{and}\:{use}\:{the}\:{ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${I}=\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{6}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 28703 by students last updated on 29/Jan/18 $$\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\:\:{solve}\:{the}\:{integration} \\ $$ Commented by abdo imad last updated on 29/Jan/18 $${let}\:{put}\:{I}=\:\int\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}}…
Question Number 28702 by students last updated on 29/Jan/18 $${solve}\:{integration}\:\:\frac{\mathrm{1}}{\:\sqrt{\left({x}−\alpha\right)\left(\beta−{x}\right)}}\:\:. \\ $$ Commented by abdo imad last updated on 29/Jan/18 $${let}\:{use}\:{the}\:{ch}.\:{x}=\:\frac{\alpha−\beta}{\mathrm{2}}{t}\:+\frac{\alpha+\beta}{\mathrm{2}}{so} \\ $$$${x}−\alpha=\:\frac{\alpha−\beta}{\mathrm{2}}{t}\:+\frac{\alpha+\beta\:−\mathrm{2}\alpha}{\mathrm{2}}=\frac{\alpha−\beta}{\mathrm{2}}\:\left({t}−\mathrm{1}\right)\:{and}\: \\ $$$$\beta−{x}=\beta−\frac{\alpha+\beta}{\mathrm{2}}\:−\frac{\alpha−\beta}{\mathrm{2}}{t}\:=\frac{\beta−\alpha}{\mathrm{2}}\:−\frac{\alpha−\beta}{\mathrm{2}}{t}…
Question Number 28701 by students last updated on 29/Jan/18 $${integration}\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$ Commented by abdo imad last updated on 29/Jan/18 $$\int\:\:\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:\int\:\frac{{x}^{\mathrm{3}}…
Question Number 159768 by cortano last updated on 21/Nov/21 Commented by Tony6400 last updated on 21/Nov/21 $$\int_{\precsim\mathrm{3}} ^{\mathrm{4}} \left[\mathrm{6}+\frac{{x}}{\mathrm{2}}\precsim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]{dx} \\ $$$$=\left[\mathrm{6}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\precsim\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right]_{\precsim\mathrm{3}} ^{\mathrm{4}}…