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Question Number 22623 Answers: 1 Comments: 0

How is: ∫ ((x + sin(x))/(1 + cos(x))) dx = xtan((x/2)) + C

$$\mathrm{How}\:\mathrm{is}:\:\:\:\int\:\frac{\mathrm{x}\:+\:\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\mathrm{xtan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:+\:\mathrm{C} \\ $$

Question Number 22621 Answers: 0 Comments: 0

∫(dx/(((√(1+x^2 ))−x)^n ))(n≠1)=(1/2)((z^(n+1) /(n+1))+(z^(n−1) /(n−1)))+cccccc where z=?

$$\int\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)^{\mathrm{n}} }\left(\mathrm{n}\neq\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{z}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{z}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}−\mathrm{1}}\right)+\mathrm{cccccc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{where}\:\:\mathrm{z}=? \\ $$$$ \\ $$

Question Number 22618 Answers: 1 Comments: 0

If (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + ... + C_n x^n , prove that (2^2 /(1.2))C_0 + (2^3 /(2.3))C_1 + (2^4 /(3.4))C_2 + ... + (2^(n+2) /((n + 1)(n + 2)))C_n = ((3^(n+2) − 2n − 5)/((n + 1)(n + 2)))

$${If}\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:{C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} {x}^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} {x}^{\mathrm{3}} \\ $$$$+\:...\:+\:{C}_{{n}} {x}^{{n}} ,\:{prove}\:{that} \\ $$$$\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}.\mathrm{2}}{C}_{\mathrm{0}} \:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}.\mathrm{3}}{C}_{\mathrm{1}} \:+\:\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}.\mathrm{4}}{C}_{\mathrm{2}} \:+\:...\:+ \\ $$$$\frac{\mathrm{2}^{{n}+\mathrm{2}} }{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}{C}_{{n}} \:=\:\frac{\mathrm{3}^{{n}+\mathrm{2}} \:−\:\mathrm{2}{n}\:−\:\mathrm{5}}{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)} \\ $$

Question Number 22640 Answers: 0 Comments: 0

With usual notation, show that (C_0 /x) − (C_1 /(x+1)) + (C_2 /(x+2)) − .... + (−1)^n (C_n /(x+n))= ((n!)/(x(x + 1)(x + 2)....(x + n)))

$${With}\:{usual}\:{notation},\:{show}\:{that} \\ $$$$\frac{{C}_{\mathrm{0}} }{{x}}\:−\:\frac{{C}_{\mathrm{1}} }{{x}+\mathrm{1}}\:+\:\frac{{C}_{\mathrm{2}} }{{x}+\mathrm{2}}\:−\:....\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{{C}_{{n}} }{{x}+{n}}= \\ $$$$\frac{{n}!}{{x}\left({x}\:+\:\mathrm{1}\right)\left({x}\:+\:\mathrm{2}\right)....\left({x}\:+\:{n}\right)} \\ $$

Question Number 23796 Answers: 1 Comments: 0

∫((2sinx+3cosx)/(3sinx+4cosx)) dx

$$\int\frac{\mathrm{2sinx}+\mathrm{3cosx}}{\mathrm{3sinx}+\mathrm{4cosx}}\:\mathrm{dx} \\ $$

Question Number 23793 Answers: 1 Comments: 1

Question Number 22598 Answers: 0 Comments: 4

Wow guys. You people are amazing.Its my pleasure to be a part of this forum. :)

$${Wow}\:{guys}.\:{You}\:{people}\:{are} \\ $$$${amazing}.{Its}\:{my}\:{pleasure}\:{to} \\ $$$$\left.{be}\:{a}\:{part}\:{of}\:{this}\:{forum}.\:\::\right) \\ $$$$\: \\ $$

Question Number 22594 Answers: 1 Comments: 0

ΔH_f ^o of water is −285.8 kJ mol^(−1) . If enthalpy of neutralization of monoacid strong base is −57.3 kJ mol^(−1) , ΔH_f ^o of OH^− ion will be

$$\Delta\mathrm{H}_{\mathrm{f}} ^{\mathrm{o}} \:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:−\mathrm{285}.\mathrm{8}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} .\:\mathrm{If} \\ $$$$\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{neutralization}\:\mathrm{of}\:\mathrm{monoacid} \\ $$$$\mathrm{strong}\:\mathrm{base}\:\mathrm{is}\:−\mathrm{57}.\mathrm{3}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} ,\:\Delta\mathrm{H}_{\mathrm{f}} ^{\mathrm{o}} \:\mathrm{of} \\ $$$$\mathrm{OH}^{−} \:\mathrm{ion}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 22599 Answers: 0 Comments: 5

Question Number 22612 Answers: 2 Comments: 0

In the binomial expasion of (a − b)^5 , the sum of 2^(nd) and 3^(rd) term is zero, then (a/b) is

$${In}\:{the}\:{binomial}\:{expasion}\:{of}\:\left({a}\:−\:{b}\right)^{\mathrm{5}} , \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term}\:{is}\:{zero}, \\ $$$${then}\:\frac{{a}}{{b}}\:{is} \\ $$

Question Number 22582 Answers: 1 Comments: 2

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2 m/s^2 . The force of interaction (in newton) between the top link and the link immediately below it will be : Take g = 10 m/s^2 .

$$\mathrm{A}\:\mathrm{chain}\:\mathrm{consisting}\:\mathrm{of}\:\mathrm{5}\:\mathrm{links}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{0}.\mathrm{1}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{The} \\ $$$$\mathrm{force}\:\mathrm{of}\:\mathrm{interaction}\:\left(\mathrm{in}\:\mathrm{newton}\right)\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{top}\:\mathrm{link}\:\mathrm{and}\:\mathrm{the}\:\mathrm{link}\:\mathrm{immediately} \\ $$$$\mathrm{below}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:: \\ $$$$\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} . \\ $$

Question Number 22576 Answers: 1 Comments: 1

Question Number 22574 Answers: 1 Comments: 1

The system is initially at origin and is moving with a velocity (3 m/s) k^∧ . A force (120t newton) i^∧ acts on mass m_2 , where t is time in seconds. The man throws a light ball, at the instant when m_1 starts slipping on m_2 , with a velocity 5 m/s vertically up w.r.t himself. Taking the masses of blocks and man as 60 kg each and assuming that the man never slips on m_2 , find (a) The time at which man throws the ball and (b) The coordinates of the point where ball lands. Neglect the dimensions of the system. (g = 10 m/s^2 )

$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\left(\mathrm{3}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{k}}.\:\mathrm{A} \\ $$$$\mathrm{force}\:\left(\mathrm{120}{t}\:\mathrm{newton}\right)\:\overset{\wedge} {{i}}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{mass}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{in}\:\mathrm{seconds}. \\ $$$$\mathrm{The}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{a}\:\mathrm{light}\:\mathrm{ball},\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{instant}\:\mathrm{when}\:{m}_{\mathrm{1}} \:\mathrm{starts}\:\mathrm{slipping}\:\mathrm{on}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{m}/\mathrm{s}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{w}.\mathrm{r}.\mathrm{t} \\ $$$$\mathrm{himself}.\:\mathrm{Taking}\:\mathrm{the}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{blocks} \\ $$$$\mathrm{and}\:\mathrm{man}\:\mathrm{as}\:\mathrm{60}\:\mathrm{kg}\:\mathrm{each}\:\mathrm{and}\:\mathrm{assuming} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{man}\:\mathrm{never}\:\mathrm{slips}\:\mathrm{on}\:{m}_{\mathrm{2}} ,\:\mathrm{find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{at}\:\mathrm{which}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{and} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where} \\ $$$$\mathrm{ball}\:\mathrm{lands}.\:\mathrm{Neglect}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{system}.\:\left({g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$

Question Number 22562 Answers: 0 Comments: 0

solve∫_0 ^1 ln(x)dx/x^2 −x−1 dx

$${solve}\overset{\mathrm{1}} {\int}_{\mathrm{0}} {ln}\left({x}\right){dx}/{x}^{\mathrm{2}} −{x}−\mathrm{1}\:{dx} \\ $$

Question Number 22561 Answers: 1 Comments: 0

Solve for all integer x,y ∈Z x^2 +y^2 =19

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{{x}},\boldsymbol{{y}}\:\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{19} \\ $$

Question Number 22547 Answers: 1 Comments: 0

If α = (5/(2!3)) + ((5.7)/(3!3^2 )) + ((5.7.9)/(4!3^3 )) ,... then find the value of α^2 + 4α.

$$\mathrm{If}\:\alpha\:=\:\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}\:+\:\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}.\mathrm{7}.\mathrm{9}}{\mathrm{4}!\mathrm{3}^{\mathrm{3}} }\:,...\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{2}} \:+\:\mathrm{4}\alpha. \\ $$

Question Number 22545 Answers: 2 Comments: 0

∫(x^(1/2) /(x^(1/2) −x^(1/3) ))dx=