Question Number 141532 by sarkor last updated on 20/May/21 Answered by qaz last updated on 20/May/21 $$\int\frac{\mathrm{3sin}\:{x}−\mathrm{2cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx} \\ $$$$=\int\frac{\mathrm{6sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}−\mathrm{4cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}}{\mathrm{2cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$=\int\left(\mathrm{3tan}\:\frac{{x}}{\mathrm{2}}−\mathrm{2}+\mathrm{sec}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right){dx} \\…
Question Number 141535 by mohammad17 last updated on 20/May/21 $${how}\:{can}\:{solve}\:{this}\:{with}\:{by}\:{steps}\:{please} \\ $$$$\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{5}\right)}\:{i}\:{want}\:{solution}\:{step}\:{by}\:{step} \\ $$$${because}\:{i}\:{understand}\:{this}? \\ $$$$ \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 141534 by sarkor last updated on 20/May/21 Answered by som(math1967) last updated on 20/May/21 $$\int\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{xsin}}^{\mathrm{8}} \boldsymbol{{xcosxdx}} \\ $$$$=\int\left(\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}\right)\boldsymbol{{sin}}^{\mathrm{8}} \boldsymbol{{xd}}\left(\boldsymbol{{sinx}}\right) \\ $$$$=\int\boldsymbol{{sin}}^{\mathrm{8}}…
Question Number 141529 by sarkor last updated on 20/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141528 by sarkor last updated on 20/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10458 by paonky last updated on 10/Feb/17 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{sin}{x}\:+\:\mathrm{cos}{x}\:+\:\mathrm{sin}{x}\mathrm{cos}{x} \\ $$ Answered by mrW1 last updated on 10/Feb/17 $$\mathrm{sin}{x}\:+\mathrm{cos}\:{x} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{sin}\:{x}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{cos}\:{x}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\…
Question Number 141531 by sarkor last updated on 20/May/21 Commented by mohammad17 last updated on 20/May/21 $${let}\:{z}^{\mathrm{6}} ={x}+\mathrm{1}\Rightarrow\mathrm{6}{z}^{\mathrm{5}} {dz}={dx} \\ $$$$ \\ $$$$\int\:\:\frac{\mathrm{6}{z}^{\mathrm{3}} \left({z}^{\mathrm{4}} +{z}\right)}{\left({z}+\mathrm{1}\right)}{dz}=\mathrm{6}\int\:\:\frac{{z}^{\mathrm{7}}…
Question Number 141530 by sarkor last updated on 20/May/21 Commented by mohammad17 last updated on 20/May/21 $$=\int\:\frac{{x}^{\mathrm{2}} −\mathrm{9}+\mathrm{9}}{\:\sqrt{{x}−\mathrm{3}}}{dx}=\int\:\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)}{\:\sqrt{{x}−\mathrm{3}}}{dx}+\int\:\frac{\mathrm{9}}{\:\sqrt{{x}−\mathrm{3}}}{dx} \\ $$$$ \\ $$$$=\int\left(\sqrt{{x}−\mathrm{3}}\right)\left({x}+\mathrm{3}\right){dx}+\int\mathrm{9}\left({x}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$…
Question Number 141525 by hgrocks last updated on 20/May/21 $$\mathrm{Let}\:<\mathrm{x}_{\mathrm{n}} >\:\mathrm{be}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{k}}\left(\mathrm{x}_{\mathrm{n}} +\frac{\mathrm{k}}{\mathrm{x}_{\mathrm{n}} }\right)\:\forall\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{Show}\:\mathrm{that}\:<\mathrm{x}_{\mathrm{n}} >\:\mathrm{converges}\:\mathrm{to}\:\sqrt{\frac{\mathrm{k}}{\mathrm{k}−\mathrm{1}}} \\ $$$$\mathrm{x}_{\mathrm{1}} >\mathrm{0}\:,\:\mathrm{k}>\mathrm{1} \\ $$ Answered…
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