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IntegrationQuestion and Answers: Page 22
Question Number 187095 Answers: 2 Comments: 1
$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }}\:\:+\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }}\:\:\boldsymbol{{dx}}\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 187050 Answers: 0 Comments: 1
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\:\frac{\boldsymbol{{cos}}^{−\mathrm{1}} \boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\:\boldsymbol{{x}}}{\boldsymbol{{Ln}}\:\boldsymbol{{x}}}\:\:\boldsymbol{{dx}}\:=??\:\:\: \\ $$$$ \\ $$
Question Number 186921 Answers: 0 Comments: 0
Question Number 186911 Answers: 1 Comments: 0
$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{a}^{{x}} +{a}^{\frac{{x}}{\mathrm{2}}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\left[\mathrm{ln}\:\mathrm{3}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$
Question Number 186910 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{{a}} \sqrt{\frac{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{a}}{\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}}}\:{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:{a}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Question Number 186973 Answers: 0 Comments: 1
$$\int\mathrm{5}{x} \\ $$
Question Number 186856 Answers: 0 Comments: 0
$$\int\:\frac{\sqrt{{x}+\mathrm{3}}+\sqrt{{x}+\mathrm{2}}}{\:\sqrt{{x}+\mathrm{3}}+\sqrt{{x}−\mathrm{2}}}\:{dx}\:=? \\ $$
Question Number 186840 Answers: 1 Comments: 0
$$\:\:\underset{\mathrm{9}} {\overset{\:\mathrm{16}} {\int}}\:\frac{\sqrt{\mathrm{4}−\sqrt{{x}}}}{{x}}\:{dx}\:=? \\ $$
Question Number 186800 Answers: 0 Comments: 0
$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty\:} \frac{\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\alpha{x}}\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\beta{x}}}{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{log}\left\{\frac{\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }{\boldsymbol{\alpha}^{\boldsymbol{\alpha}} \boldsymbol{\beta}^{\boldsymbol{\beta}} }\right\}\: \\ $$
Question Number 186771 Answers: 2 Comments: 0
$$ \\ $$$$\:\:\:\:\mathrm{Q}\::\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral}.\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\frac{\:\pi}{\:\mathrm{2}}} \:\frac{\:\:\mathrm{1}}{\:\mathrm{1}\:+\:\mathrm{sin}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:+\:\mathrm{cos}^{\:\mathrm{4}} \:\left(\:{x}\:\right)\:}\:\mathrm{d}{x}\:=\:\:?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 186780 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\int\:\:\:\frac{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\:\boldsymbol{{x}}}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{{x}}}\:\:\boldsymbol{{dx}}\:\: \\ $$$$ \\ $$
Question Number 186736 Answers: 1 Comments: 0
Question Number 186735 Answers: 1 Comments: 0
Question Number 186726 Answers: 2 Comments: 0
Question Number 186637 Answers: 2 Comments: 0
Question Number 186531 Answers: 0 Comments: 0
$$ \\ $$$$\:\:\mathbb{Q}.\boldsymbol{{use}}\:\boldsymbol{{the}}\:\boldsymbol{{parseval}}\:\boldsymbol{{relation}}\:\boldsymbol{{of}}\:\boldsymbol{{hankel}}\:\boldsymbol{{transfrom}}\:\boldsymbol{{to}}\:\boldsymbol{{evaluate}}\:\boldsymbol{{the}}\:\boldsymbol{{Integral}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\infty} {\int}_{\mathrm{0}} \:\:\frac{\boldsymbol{{J}}_{\boldsymbol{\gamma}+\mathrm{1}} \left(\boldsymbol{{ar}}\right)\boldsymbol{{J}}_{\boldsymbol{\gamma}+\mathrm{1}} \left(\boldsymbol{{br}}\right)}{\boldsymbol{{r}}}\:,\:\:\boldsymbol{{for}}\:\boldsymbol{\gamma}>−\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:\mathrm{0}<\boldsymbol{{a}}<\boldsymbol{{b}} \\ $$$$\:\:\:\boldsymbol{{where}}\:\boldsymbol{{J}}_{\boldsymbol{{n}}} \left(\boldsymbol{{x}}\right)\:\boldsymbol{{are}}\:\boldsymbol{{bessel}}\:\boldsymbol{{funtions}}. \\ $$$$ \\ $$
Question Number 186527 Answers: 2 Comments: 0
Question Number 186476 Answers: 0 Comments: 0
Question Number 186347 Answers: 1 Comments: 1
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\:\:\:\:\frac{\boldsymbol{{x}}^{\mathrm{5}} \:−\:\:\mathrm{1}\:\:+\:\:\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{4}} \:\:+\:\:\boldsymbol{{x}}\:\:−\mathrm{2}}\:\:\:\:\boldsymbol{{dx}}\:\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 186346 Answers: 0 Comments: 0
$${if}\:{S}_{{a}} =\mathrm{cos}\left({a}\right)+\mathrm{sin}\left({x}+{a}\right) \\ $$$${then}\:\int\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }−\frac{{x}+{S}_{\mathrm{1}} }{{x}−{S}_{\mathrm{3}} }=? \\ $$
Question Number 186321 Answers: 3 Comments: 0
Question Number 186310 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\frac{\int\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{1}/\mathrm{2}} \boldsymbol{{dx}}\:−\:\mathrm{3}\int\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)^{−\mathrm{1}/\mathrm{2}} \:\boldsymbol{{dx}}}{\int\:\:\frac{\boldsymbol{{x}}\left[\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)−\mathrm{3}\right]}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\:\:}}\:\boldsymbol{{dx}}}\:=??\:\:\:\: \\ $$$$ \\ $$
Question Number 186306 Answers: 0 Comments: 2
$$\mathrm{Evaluate}\:\int\frac{\mathrm{ln}\left(\mathrm{sin}\:{x}\right)}{\mathrm{ln}\left(\mathrm{tan}\:{x}\right)+\mathrm{1}}\:{dx} \\ $$
Question Number 186246 Answers: 1 Comments: 0
Question Number 186352 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\:\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \:\left(\boldsymbol{{x}}\right)\:+\:\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} }\:\:\boldsymbol{{dx}}\:\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 186214 Answers: 0 Comments: 12
$${if}\:{S}_{{a}} =\mathrm{cos}\left({a}\right)+\mathrm{sin}\left({x}+{a}\right) \\ $$$${then}\:\int\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }−\frac{{x}+{S}_{\mathrm{1}} }{{x}−{S}_{\mathrm{3}} }{dx}=? \\ $$
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