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If ∫ ((tan x)/(1+tan x+tan^2 x)) dx = x−(k/( (√A))) tan^(−1) (((k tan x+1)/( (√A))))+C where C is constant of integration. then the ordered pair (k,A) is equal to

$$\mathrm{If}\:\int\:\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{tan}\:\mathrm{x}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=\:\mathrm{x}−\frac{{k}}{\:\sqrt{{A}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{k}\:\mathrm{tan}\:{x}+\mathrm{1}}{\:\sqrt{{A}}}\right)+\mathrm{C} \\ $$$$\mathrm{where}\:\mathrm{C}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{of}\:\mathrm{integration}. \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{ordered}\:\mathrm{pair}\:\left({k},\mathrm{A}\right)\:\mathrm{is}\: \\ $$$$\mathrm{equal}\:\mathrm{to}\: \\ $$

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Question Number 132987 Answers: 2 Comments: 0

....mathematical analysis... prove that:: 𝛗=∫_0 ^( ∞) ((sin^3 (x))/x^3 )dx=((3π)/8) ∗∗∗∗..........

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{mathematical}\:\:{analysis}... \\ $$$$\:{prove}\:\:{that}::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\ast\ast\ast\ast.......... \\ $$$$ \\ $$

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(∫_0 ^(π/2) (√(sin(x)))dx)^2 +(∫_0 ^(π/2) (√(cos(x)))dx)^2 <8((√2)−1)

$$\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}\right)^{\mathrm{2}} +\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left({x}\right)}{dx}\right)^{\mathrm{2}} <\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$

Question Number 132901 Answers: 2 Comments: 0

∫_(−π/2) ^(3π/2) (sin^(−1) (∣sin x∣)+cos^(−1) (∣cos x∣)) dx

$$\int_{−\pi/\mathrm{2}} ^{\mathrm{3}\pi/\mathrm{2}} \:\left(\mathrm{sin}^{−\mathrm{1}} \left(\mid\mathrm{sin}\:\mathrm{x}\mid\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mid\mathrm{cos}\:\mathrm{x}\mid\right)\right)\:\mathrm{dx} \\ $$

Question Number 132877 Answers: 1 Comments: 0

.... nice calculus... prove that : 𝛗=∫_0 ^( ∞) ((sin(x).log^2 (x))/x) =(π/(24))(12γ^( 2) +π^2 ) .....

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....\:{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right).{log}^{\mathrm{2}} \left({x}\right)}{{x}} \\ $$$$\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{24}}\left(\mathrm{12}\gamma^{\:\mathrm{2}} +\pi^{\mathrm{2}} \right)\:..... \\ $$

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∫(x−1)^(x+1) dx

$$\int\left({x}−\mathrm{1}\right)^{{x}+\mathrm{1}} \mathrm{d}{x} \\ $$

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

$$\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \left(\sqrt{\mathrm{sin}\:\left({x}\right)}+\sqrt{\mathrm{cos}\:\left({x}\right)}\right){dx} \\ $$

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∫_0 ^∞ ((log x)/(1+x^2 +x^4 )) dx=...?

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }\:{dx}=...? \\ $$

Question Number 132729 Answers: 3 Comments: 0

....advanced calculus... evaluate : 𝛗=∫_0 ^( ∞) xe^(−2x) ln(x)dx=???

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:\:\:{calculus}... \\ $$$$\:\:\:\:{evaluate}\:: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} {xe}^{−\mathrm{2}{x}} {ln}\left({x}\right){dx}=??? \\ $$$$ \\ $$

Question Number 132708 Answers: 2 Comments: 0

∫_(−∞) ^∞ ((x^2 cos (px+q))/(x^2 +(p+q)^2 ))dx

$$\int_{−\infty} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:\left({px}+{q}\right)}{{x}^{\mathrm{2}} +\left({p}+{q}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 132693 Answers: 2 Comments: 0

I=∫ (dx/(x(x^2 +1)^3 ))

$$\mathrm{I}=\int\:\frac{{dx}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\: \\ $$

Question Number 132610 Answers: 2 Comments: 0

Ω=∫ ((sin^2 (x))/(1+sin^2 (x))) dx

$$\Omega=\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$

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Find the voloume bounded by z=(√(x^2 +y^2 )) and the plane y+z=3

$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{voloume}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{z}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{y}+\mathrm{z}=\mathrm{3} \\ $$

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