Question

Double Integral with Variable Substitution

score 4.1k · Answer 1 · 2017-04-09 23:09:36

4.1k

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Double Integral with Variable Substitution

Published on 09 Apr 2017 - 23:09

score 61 · Answer 2 · 2017-04-10 14:22:59

61

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Is this the proper use of U-Substitution?

Published on 10 Apr 2017 - 14:22

score 89 · Answer 3 · 2026-03-26 19:32:20

89

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Integration by part with substitution

Published on 26 Mar 2026 - 19:32

score 164 · Answer 4 · 2017-04-15 12:55:35

164

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Solve$\int\frac{x^4}{1-x^4}dx$

Published on 15 Apr 2017 - 12:55

score 119 · Answer 5 · 2017-04-15 15:19:51

119

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Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$

Published on 15 Apr 2017 - 15:19

score 43 · Answer 6 · 2017-04-16 16:31:22

43

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Integration by subtitution

Published on 16 Apr 2017 - 16:31

score 127 · Answer 7 · 2017-04-17 14:00:55

127

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How to integrate $\int^{2\pi}_{0} \frac{dx}{\mathrm{sin}^4x +\mathrm{cos}^4x}$?

Published on 17 Apr 2017 - 14:00

score 128 · Answer 8 · 2026-03-26 11:02:19

128

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Prove this inequality $(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)^3\ge 27(ab+bc+ac)^2$

Published on 26 Mar 2026 - 11:02

score 193 · Answer 9 · 2017-04-18 12:25:38

193

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How to integrate $\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$

Published on 18 Apr 2017 - 12:25

score 319 · Answer 10 · 2017-04-21 07:39:16

319

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Undoing Trigonometric Substitution

Published on 21 Apr 2017 - 7:39

score 178 · Answer 11 · 2026-03-30 10:23:41

178

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Prove that $\sum\limits_{i=1}^n\left(\frac{x_i}{x_{i+1}}\right)^{4}\geq\sum\limits_{i=1}^n\frac{x_i}{x_{i+4}}$

Published on 30 Mar 2026 - 10:23

score 374 · Answer 12 · 2017-04-21 16:26:05

374

Views

Please explain why this equation doesn't work with this form of U-substitution?

Published on 21 Apr 2017 - 16:26

score 5.8k · Answer 13 · 2017-04-22 06:36:31

5.8k

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Solving integral given the value of another integral

Published on 22 Apr 2017 - 6:36

score 104 · Answer 14 · 2017-04-22 13:43:22

104

Views

Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}} $?

Published on 22 Apr 2017 - 13:43

score 406 · Answer 15 · 2017-04-29 14:21:00

406

Views

Why can we simplify $\sqrt{1 - \sin^2 u} = \sqrt {\cos^2 u}$ to $\cos u$ instead of $|\cos u|$ when making a trigonometric substitution?

Published on 29 Apr 2017 - 14:21

List Question

Double Integral with Variable Substitution

Is this the proper use of U-Substitution?

Integration by part with substitution

Solve$\int\frac{x^4}{1-x^4}dx$

Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$

Integration by subtitution

How to integrate $\int^{2\pi}_{0} \frac{dx}{\mathrm{sin}^4x +\mathrm{cos}^4x}$?

Prove this inequality $(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)^3\ge 27(ab+bc+ac)^2$

How to integrate $\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$

Undoing Trigonometric Substitution

Prove that $\sum\limits_{i=1}^n\left(\frac{x_i}{x_{i+1}}\right)^{4}\geq\sum\limits_{i=1}^n\frac{x_i}{x_{i+4}}$

Please explain why this equation doesn't work with this form of U-substitution?

Solving integral given the value of another integral

Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}} $?

Why can we simplify $\sqrt{1 - \sin^2 u} = \sqrt {\cos^2 u}$ to $\cos u$ instead of $|\cos u|$ when making a trigonometric substitution?

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