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Question Number 71918 Answers: 2 Comments: 2
$$\mathrm{If}\:\mathrm{Cos}\theta=\frac{\mathrm{x}\:\mathrm{cos}\beta\:−\:\mathrm{y}}{\mathrm{x}\:−\:\mathrm{y}\:\mathrm{cos}\beta}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{tan}\frac{\theta}{\mathrm{2}}\:=\sqrt{\frac{\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}}\:}\:\mathrm{tan}\frac{\beta}{\mathrm{2}} \\ $$
Question Number 69667 Answers: 1 Comments: 0
Question Number 69665 Answers: 1 Comments: 1
Question Number 69662 Answers: 1 Comments: 0
$${prove}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\:=\:\mathrm{2} \\ $$
Question Number 69645 Answers: 1 Comments: 0
Question Number 69644 Answers: 2 Comments: 0
Question Number 69643 Answers: 1 Comments: 0
Question Number 69641 Answers: 0 Comments: 0
$$\boldsymbol{{Solve}}\:\:\boldsymbol{{arctg}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\:+\:\frac{\mathrm{1}}{\boldsymbol{{arcctg}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$
Question Number 69637 Answers: 1 Comments: 0
$$...\mathrm{now}\:\mathrm{try}\:\mathrm{this}\:\mathrm{one}: \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{2}} −{x}^{\mathrm{1}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{6}} }= \\ $$
Question Number 69623 Answers: 1 Comments: 0
$$\int\frac{\mathrm{1}}{\sqrt{{x}}\:+\:\sqrt[{\mathrm{3}}]{{x}}}\:{dx} \\ $$
Question Number 69616 Answers: 0 Comments: 4
Question Number 69606 Answers: 0 Comments: 0
Question Number 69603 Answers: 1 Comments: 0
$$\int\:{x}^{\mathrm{3}} {arcsinxdx} \\ $$
Question Number 69597 Answers: 1 Comments: 1
Question Number 69594 Answers: 1 Comments: 0
Question Number 69593 Answers: 1 Comments: 2
Question Number 69576 Answers: 1 Comments: 1
$$\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$
Question Number 69574 Answers: 1 Comments: 0
Question Number 69573 Answers: 0 Comments: 1
Question Number 69572 Answers: 1 Comments: 0
Question Number 69571 Answers: 1 Comments: 0
Question Number 69570 Answers: 0 Comments: 0
Question Number 69569 Answers: 2 Comments: 0
Question Number 69568 Answers: 1 Comments: 3
Question Number 69567 Answers: 0 Comments: 2
Question Number 69566 Answers: 1 Comments: 0
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