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Question Number 83127 Answers: 1 Comments: 0
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{x}\sqrt{\mathrm{3}}\:+\mathrm{y}\: \\ $$$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\:\mathrm{3}\:? \\ $$
Question Number 83123 Answers: 0 Comments: 6
$$\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}\right)}\:{dx} \\ $$
Question Number 83115 Answers: 1 Comments: 1
$$\int\mathrm{cos}\:{xe}^{\mathrm{sin}\:{x}} {dx} \\ $$
Question Number 83110 Answers: 0 Comments: 10
$${bounded}\:{by}\:{the}\:{curve}\:{y}=\sqrt{\mathrm{4}-{x}}\:{y}=\mathrm{0}\:{y}=\mathrm{1} \\ $$
Question Number 83109 Answers: 0 Comments: 1
$$\int_{\mathrm{1}/\boldsymbol{{e}}} ^{{e}} \frac{\boldsymbol{{dt}}}{\boldsymbol{{t}}} \\ $$
Question Number 83108 Answers: 1 Comments: 0
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cos}\left({nx}\right)}{{cos}^{{n}} \left({x}\right)}\:{dx}\:=\mathrm{2}^{{n}} \left[\frac{\pi}{\mathrm{8}}−\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\mathrm{2}{k}\left(\sqrt{\mathrm{2}}\right)^{{k}} }\right]\:{n}\in{N}^{\ast} \\ $$
Question Number 83104 Answers: 1 Comments: 0
$$\int\frac{{e}^{{x}} {dx}}{\mathrm{3}+{e}^{{x}} } \\ $$
Question Number 83102 Answers: 0 Comments: 3
Question Number 83096 Answers: 0 Comments: 1
$$\int\mathrm{tan}\:{x}^{\mathrm{4}} {dx} \\ $$
Question Number 83095 Answers: 0 Comments: 0
$$\int\mathrm{cosec}\:{x}^{\mathrm{5}} {dx} \\ $$
Question Number 83094 Answers: 0 Comments: 0
Question Number 83093 Answers: 0 Comments: 1
Question Number 83092 Answers: 0 Comments: 0
Question Number 83085 Answers: 1 Comments: 2
$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} } \\ $$
Question Number 83078 Answers: 0 Comments: 1
$$\mathrm{If}\:\:\:{I}_{\mathrm{1}} =\underset{{e}} {\overset{{e}^{\mathrm{2}} } {\int}}\:\frac{{dx}}{\mathrm{log}\:{x}}\:\:\mathrm{and}\:\:{I}_{\mathrm{2}} =\:\underset{\:\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{{e}^{{x}} }{{x}}\:{dx},\:\mathrm{then} \\ $$
Question Number 83077 Answers: 1 Comments: 0
Question Number 83075 Answers: 1 Comments: 1
$$\underset{−\mathrm{1}\:\:} {\overset{\mathrm{3}} {\int}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\right){dx}\:= \\ $$
Question Number 83074 Answers: 0 Comments: 2
$${find}\:{all}\:{function}\:\:{satisfying}\:\:\forall\:{x}\in\mathbb{R}\backslash\left\{{k}\pi\:,\:\:{k}\in\mathbb{Z}\right\} \\ $$$${f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}=\frac{{x}}{{sin}\left(\pi{x}\right)} \\ $$
Question Number 83064 Answers: 2 Comments: 0
$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left({nx}\right)}{{sin}\left({x}\right)}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$${n}\:{is}\:{posative}\:{odd}\:{number} \\ $$
Question Number 83063 Answers: 1 Comments: 3
Question Number 83050 Answers: 0 Comments: 1
Question Number 83049 Answers: 0 Comments: 0
Question Number 83042 Answers: 0 Comments: 1
$${let}\:\:{a},{b}\:{two}\:{positive}\:{reals}\:{such}\:{as}\:\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$${Explicit}\:\:\:{f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{du}}{\sqrt{{a}+{bsin}^{\mathrm{2}} {u}}}\: \\ $$
Question Number 83037 Answers: 0 Comments: 3
$$\mathrm{If}\:\mathrm{m}=\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{sin}\theta}\:,\:\:\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{m}}=\:\frac{\mathrm{1}+\mathrm{sin}\theta}{\mathrm{sin}\theta} \\ $$
Question Number 83036 Answers: 0 Comments: 3
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{cos}^{\mathrm{4}} \theta−\mathrm{sin}^{\mathrm{4}} \theta=\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta \\ $$
Question Number 83035 Answers: 1 Comments: 1
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