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AllQuestion and Answers: Page 1406

Question Number 64119    Answers: 3   Comments: 1

Question Number 64111    Answers: 0   Comments: 0

Question Number 64112    Answers: 2   Comments: 1

Question Number 64106    Answers: 0   Comments: 0

∫tan(1/x)dx

$$\int{tan}\left(\mathrm{1}/{x}\right){dx} \\ $$

Question Number 64105    Answers: 0   Comments: 0

∫tan(1/x)dx

$$\int{tan}\left(\mathrm{1}/{x}\right){dx} \\ $$

Question Number 64104    Answers: 0   Comments: 0

∫tan(1/x)dx

$$\int{tan}\left(\mathrm{1}/{x}\right){dx} \\ $$

Question Number 64103    Answers: 0   Comments: 0

∫tan(1/x)dx

$$\int{tan}\left(\mathrm{1}/{x}\right){dx} \\ $$

Question Number 64102    Answers: 0   Comments: 0

∫tan(1/x)dx

$$\int{tan}\left(\mathrm{1}/{x}\right){dx} \\ $$

Question Number 64101    Answers: 0   Comments: 1

why the first ionisation(△H_(i1) )energy of Oxygen smaller than the second? A. Due to the nuclear charge B. Due to the distance of the electron from the nucleus C. Due to the effect of spin−pair repulsion D. Due to a shielding effect.

$${why}\:{the}\:{first}\:{ionisation}\left(\bigtriangleup{H}_{{i}\mathrm{1}} \right){energy}\:{of}\:{Oxygen}\:{smaller} \\ $$$${than}\:{the}\:{second}? \\ $$$${A}.\:{Due}\:{to}\:{the}\:{nuclear}\:{charge} \\ $$$${B}.\:{Due}\:{to}\:{the}\:{distance}\:{of}\:{the}\:{electron}\:{from}\:{the}\:{nucleus} \\ $$$${C}.\:{Due}\:{to}\:{the}\:{effect}\:{of}\:{spin}−{pair}\:{repulsion} \\ $$$${D}.\:{Due}\:{to}\:{a}\:{shielding}\:{effect}. \\ $$

Question Number 64097    Answers: 1   Comments: 1

Question Number 64092    Answers: 1   Comments: 0

prove that u=mgh

$${prove}\:{that}\: \\ $$$${u}={mgh} \\ $$

Question Number 64085    Answers: 0   Comments: 0

please just read equation of a line and a plane in vectors. i don′t understand (r−a)×b=0 ??

$${please}\:{just}\:{read}\:{equation}\:{of}\:{a}\:{line}\:{and}\:{a}\:{plane}\:{in}\:{vectors}. \\ $$$${i}\:{don}'{t}\:{understand}\: \\ $$$$\:\:\left(\boldsymbol{{r}}−\boldsymbol{{a}}\right)×\boldsymbol{{b}}=\mathrm{0}\:\:?? \\ $$

Question Number 64086    Answers: 0   Comments: 5

if 3x + 5y = 1 use Bezout′s identity to find the value of x and y

$${if}\:\:\:\mathrm{3}{x}\:+\:\mathrm{5}{y}\:=\:\mathrm{1} \\ $$$${use}\:{Bezout}'{s}\:{identity}\:{to}\:{find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y} \\ $$

Question Number 64081    Answers: 0   Comments: 0

∫e^(sec x) ∙ sec^3 x(sin^2 x+cos x+sin x+sin x cos x) dx=

$$\int{e}^{\mathrm{sec}\:{x}} \centerdot\:\mathrm{sec}^{\mathrm{3}} {x}\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}\:{x}+\mathrm{sin}\:{x}+\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right)\:{dx}= \\ $$

Question Number 64080    Answers: 1   Comments: 2

∫((sin x−cos x)/(√(1−sin 2x))) e^(sin x) cos x dx =

$$\int\frac{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}}\:{e}^{\mathrm{sin}\:{x}} \mathrm{cos}\:{x}\:{dx}\:= \\ $$

Question Number 64079    Answers: 0   Comments: 0

∫_( 0) ^π ((x sin x)/(1+cos^2 x)) dx =

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

Question Number 64078    Answers: 0   Comments: 3

Question Number 64077    Answers: 0   Comments: 0

Question Number 64074    Answers: 0   Comments: 0

by using laplase transform find laplase(tan(t))

$${by}\:{using}\:{laplase}\:{transform}\:{find}\:{laplase}\left({tan}\left({t}\right)\right) \\ $$

Question Number 64068    Answers: 0   Comments: 1

calculate ∫_0 ^π ((tsint)/(3+sin^2 t)) dt

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{tsint}}{\mathrm{3}+{sin}^{\mathrm{2}} {t}}\:{dt}\: \\ $$

Question Number 64066    Answers: 1   Comments: 0

let α ,β and λ the roots of x^3 +2x−1 =0 find the value of A =α^2 +β^2 +λ^2 and B =α^3 +β^3 +λ^3 .

$${let}\:\alpha\:,\beta\:{and}\:\lambda\:{the}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\:=\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \:{and}\:\:{B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \:+\lambda^{\mathrm{3}} \:. \\ $$

Question Number 64065    Answers: 0   Comments: 3

calculate ∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 64061    Answers: 1   Comments: 0

(2x+3)^2 +25/(x+3)^2 =(√2)

$$\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{25}/\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$

Question Number 64060    Answers: 0   Comments: 1

(2x+3)^2 +25/(x+3)^2 =(√2)

$$\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{25}/\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$

Question Number 64047    Answers: 0   Comments: 3

Question Number 64037    Answers: 0   Comments: 1

reduction formulas for n∈N, some n>0, some n>1 ∫sin^n x dx=−(1/n)cos x sin^(n−1) x +((n−1)/n)∫sin^(n−2) x dx ∫cos^n x dx=(1/n)sin x cos^(n−1) x +((n−1)/n)∫cos^(n−2) x dx ∫tan^n x dx=(1/(n−1))tan^(n−1) x −∫tan^(n−2) x dx ∫sec^n x dx=(1/(n−1))tan x sec^(n−2) x +((n−2)/(n−1))∫sec^(n−2) x dx ∫csc^n x dx=−(1/(n−1))cot x csc^(n−2) x +((n−2)/(n−1))∫csc^(n−2) x dx ∫cot^n x dx=−(1/(n−1))cot^(n−1) x −∫cot^(n−2) x dx

$$\mathrm{reduction}\:\mathrm{formulas}\:\mathrm{for}\:{n}\in\mathbb{N},\:\mathrm{some}\:{n}>\mathrm{0},\:\mathrm{some}\:{n}>\mathrm{1} \\ $$$$ \\ $$$$\int\mathrm{sin}^{{n}} \:{x}\:{dx}=−\frac{\mathrm{1}}{{n}}\mathrm{cos}\:{x}\:\mathrm{sin}^{{n}−\mathrm{1}} \:{x}\:+\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{sin}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{cos}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}}\mathrm{sin}\:{x}\:\mathrm{cos}^{{n}−\mathrm{1}} \:{x}\:+\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{cos}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{tan}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{tan}^{{n}−\mathrm{1}} \:{x}\:−\int\mathrm{tan}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{sec}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{tan}\:{x}\:\mathrm{sec}^{{n}−\mathrm{2}} \:{x}\:+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{sec}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{csc}^{{n}} \:{x}\:{dx}=−\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{cot}\:{x}\:\mathrm{csc}^{{n}−\mathrm{2}} \:{x}\:+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{csc}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{cot}^{{n}} \:{x}\:{dx}=−\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{cot}^{{n}−\mathrm{1}} \:{x}\:−\int\mathrm{cot}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$

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