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IntegrationQuestion and Answers: Page 25

Question Number 183706 Answers: 1 Comments: 0

A=∫ ((1+cos^2 x)/(1+sin^4 x)) dx

$$\:\:\:\:{A}=\int\:\frac{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{4}} {x}}\:{dx} \\ $$

Question Number 183559 Answers: 1 Comments: 0

∫_1 ^∞ (dx/(x^2 +(√x))) = ?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{{x}}}\:=\:?\: \\ $$

Question Number 183531 Answers: 0 Comments: 0

Question Number 183450 Answers: 2 Comments: 0

∫_0 ^x^2 f(2t^3 −t^2 −6)dt = ln x then f(106)=?

$$\:\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:{f}\left(\mathrm{2}{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −\mathrm{6}\right){dt}\:=\:\mathrm{ln}\:{x}\: \\ $$$$\:{then}\:{f}\left(\mathrm{106}\right)=? \\ $$

Question Number 183382 Answers: 2 Comments: 4

Question Number 183320 Answers: 1 Comments: 0

∫^(π/2) _0 (dx/(cox(x/2)∙cos(x/2^2 )∙∙∙∙∙cos(x/2^n )))=?

$$\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{cox}\frac{{x}}{\mathrm{2}}\centerdot{cos}\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\centerdot\centerdot\centerdot\centerdot\centerdot{cos}\frac{{x}}{\mathrm{2}^{{n}} }}=? \\ $$

Question Number 183294 Answers: 0 Comments: 0

Question Number 183279 Answers: 1 Comments: 0

∫ ((sin x+cos x)/((tan x−cot x)^3 )) dx =?

$$\:\:\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\left(\mathrm{tan}\:{x}−\mathrm{cot}\:{x}\right)^{\mathrm{3}} }\:{dx}\:=? \\ $$

Question Number 183243 Answers: 1 Comments: 0

∫ (((√(sin x)) −(√(cos x)))/( (√(sin x)) + (√(cos x)))) dx =?

$$\:\:\int\:\frac{\sqrt{\mathrm{sin}\:{x}}\:−\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}}\:+\:\sqrt{\mathrm{cos}\:{x}}}\:{dx}\:=? \\ $$

Question Number 183157 Answers: 1 Comments: 0

∫ (dx/( ((x^3 +2019))^(1/3) )) =?

$$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{2019}}}\:=? \\ $$

Question Number 183156 Answers: 1 Comments: 0

∫ ((sin 2x dx)/(sin x−sin^2 2x)) =?

$$\:\int\:\frac{\mathrm{sin}\:\mathrm{2}{x}\:{dx}}{\mathrm{sin}\:{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:=? \\ $$

Question Number 183109 Answers: 2 Comments: 0

∫_0 ^(π/2) ((√(sin x+1))/( (√(cos x+1)))) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\sqrt{\mathrm{sin}\:{x}+\mathrm{1}}}{\:\sqrt{\mathrm{cos}\:{x}+\mathrm{1}}}\:{dx}\:=?\: \\ $$

Question Number 186086 Answers: 0 Comments: 0

Question Number 183031 Answers: 1 Comments: 0

Question Number 182995 Answers: 1 Comments: 0

Question Number 182962 Answers: 2 Comments: 0

Question Number 182902 Answers: 0 Comments: 0

∫ x^x dx

$$\int\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:\:\boldsymbol{\mathrm{dx}} \\ $$

Question Number 182891 Answers: 1 Comments: 0

∫_(0 ) ^∞ e^x ((sinmx)/x)dx=?

$$\int_{\mathrm{0}\:\:} ^{\infty} {e}^{{x}} \frac{{sinmx}}{{x}}{dx}=? \\ $$

Question Number 182769 Answers: 1 Comments: 0

∫_0 ^(π/4) (dx/(sin (x−1)−1)) =?

$$\:\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{{dx}}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)−\mathrm{1}}\:=?\: \\ $$

Question Number 182648 Answers: 3 Comments: 0

∫_0 ^(π/4) ((sec^2 x)/(tan x−sec x)) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 182636 Answers: 0 Comments: 2

∫_(−1) ^3 (1/x^2 ) dx=?

$$\underset{−\mathrm{1}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{dx}=? \\ $$

Question Number 182494 Answers: 0 Comments: 4

Question Number 182368 Answers: 1 Comments: 0

find volume of region in R^3 given by 3∣x∣ + 4∣y∣ +3∣z∣ ≤12 is

$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{region}\:\mathrm{in}\:\:\mathbb{R}^{\mathrm{3}} \:\:\mathrm{given}\:\mathrm{by}\: \\ $$$$\:\:\:\:\mathrm{3}\mid\mathrm{x}\mid\:+\:\mathrm{4}\mid\mathrm{y}\mid\:+\mathrm{3}\mid\mathrm{z}\mid\:\leqslant\mathrm{12}\:\:\mathrm{is} \\ $$

Question Number 182367 Answers: 1 Comments: 0

find volume of region bounded above by z = 1+(√(1−x^2 −y^2 )) and below by z = (√(x^2 +y^2 ))

$$ \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{above}\: \\ $$$$\:\:\:\mathrm{by}\:\mathrm{z}\:=\:\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:}\:\:\mathrm{and}\:\mathrm{below} \\ $$$$\:\:\:\:\mathrm{by}\:\:\:\mathrm{z}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:}\: \\ $$

Question Number 182295 Answers: 1 Comments: 0

Question Number 182291 Answers: 0 Comments: 0

I_2 =∫_0 ^∞ (((tan^(−1) x)/x))^2 dx=?? I_3 =∫_0 ^∞ (((tan^(−1) x)/x))^3 dx=??? I_4 =∫_0 ^∞ (((tan^(−1) x)/x))^4 dx=????

$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}\right)^{\mathrm{2}} {dx}=?? \\ $$$${I}_{\mathrm{3}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}\right)^{\mathrm{3}} {dx}=??? \\ $$$${I}_{\mathrm{4}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}\right)^{\mathrm{4}} {dx}=???? \\ $$

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