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Question Number 138691 Answers: 1 Comments: 0

∫ cos 2x (√(1+sin^2 x)) dx =?

$$\int\:\mathrm{cos}\:\mathrm{2}{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\:=? \\ $$

Question Number 138683 Answers: 1 Comments: 0

...nice .. ... ... calculus... find the value of: Θ=Σ_(n=1) ^∞ (((−1)^n sin^2 (n))/n)=? .........................

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:..\:...\:...\:{calculus}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {sin}^{\mathrm{2}} \left({n}\right)}{{n}}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......................... \\ $$

Question Number 138690 Answers: 1 Comments: 1

((x/(12)))^(log_(√3) x) =((x/(18)))^(log_(√2) x) find x

$$\left(\frac{\mathrm{x}}{\mathrm{12}}\right)^{\mathrm{log}_{\sqrt{\mathrm{3}}} \mathrm{x}} =\left(\frac{\mathrm{x}}{\mathrm{18}}\right)^{\mathrm{log}_{\sqrt{\mathrm{2}}} \mathrm{x}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 138673 Answers: 4 Comments: 3

x^2 −x=72 y^2 −y=30 x+y=4 x−y=?

$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}=\mathrm{72} \\ $$$$\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}=\mathrm{30} \\ $$$$\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{4} \\ $$$$\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}=? \\ $$

Question Number 138661 Answers: 0 Comments: 3

Question Number 138658 Answers: 1 Comments: 2

Question Number 138656 Answers: 1 Comments: 0

I=∫(dx/((px+q)(√(ax^2 +bx+c))))

$${I}=\int\frac{{dx}}{\left({px}+{q}\right)\sqrt{{ax}^{\mathrm{2}} +{bx}+{c}}} \\ $$

Question Number 138643 Answers: 2 Comments: 0

Question Number 138641 Answers: 1 Comments: 0

Question Number 138628 Answers: 1 Comments: 1

find the region in which the function f(z)=((log(z−2i))/(z^2 +1)) is analytic ? help me sir

$${find}\:{the}\:{region}\:{in}\:{which}\:{the}\:{function}\: \\ $$$$ \\ $$$${f}\left({z}\right)=\frac{{log}\left({z}−\mathrm{2}{i}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:{is}\:{analytic}\:? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\: \\ $$

Question Number 138627 Answers: 2 Comments: 1

x^2 =16^x find x

$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\mathrm{16}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 142444 Answers: 1 Comments: 0

lim_(x→(π/3)) ((tan^2 x−(√3)tanx)/(cos(x+(π/6))))=?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {{lim}}\:\frac{{tan}^{\mathrm{2}} {x}−\sqrt{\mathrm{3}}{tanx}}{{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)}=? \\ $$

Question Number 142443 Answers: 1 Comments: 0

F(x) = ∫_( x) ^( x^2 ) (1/( ln(t) )) dt Show that : ∣ F(x) ∣ ≤ ((∣ x^2 − x ∣)/(∣ ln(x) ∣)) Please

$$\:\:\:\:\:\:{F}\left({x}\right)\:=\:\int_{\:{x}} ^{\:\:{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\:\mathrm{ln}\left({t}\right)\:}\:\mathrm{d}{t} \\ $$$$\:\:\mathrm{Show}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{F}\left({x}\right)\:\mid\:\:\leqslant\:\:\frac{\mid\:{x}^{\mathrm{2}} \:−\:{x}\:\mid}{\mid\:\mathrm{ln}\left({x}\right)\:\mid} \\ $$$$ \\ $$$$\mathrm{Please} \\ $$

Question Number 138624 Answers: 0 Comments: 0

(1^π /(3!))+(2^π /(5!))+(3^π /(7!))+(4^π /(9!))+(5^π /(11!))+...

$$\frac{\mathrm{1}^{\pi} }{\mathrm{3}!}+\frac{\mathrm{2}^{\pi} }{\mathrm{5}!}+\frac{\mathrm{3}^{\pi} }{\mathrm{7}!}+\frac{\mathrm{4}^{\pi} }{\mathrm{9}!}+\frac{\mathrm{5}^{\pi} }{\mathrm{11}!}+... \\ $$

Question Number 138621 Answers: 1 Comments: 0

Question Number 138619 Answers: 0 Comments: 1

∫_0 ^∞ (e^(−x^(√3) ) −(1/((1+x^(√3) )^(√3) )))(dx/x)=(1/( (√3)))ψ((1/( (√3))))

$$\int_{\mathrm{0}} ^{\infty} \left({e}^{−{x}^{\sqrt{\mathrm{3}}} } −\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\sqrt{\mathrm{3}}} \right)^{\sqrt{\mathrm{3}}} }\right)\frac{{dx}}{{x}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\psi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$

Question Number 138613 Answers: 1 Comments: 0

Question Number 138612 Answers: 1 Comments: 0

If z+ (1/z) =2cosβ .show that z^m +(1/z^m ) =2cosmβ.

$$\mathrm{If}\:\mathrm{z}+\:\frac{\mathrm{1}}{\mathrm{z}}\:=\mathrm{2cos}\beta\:.\mathrm{show}\:\mathrm{that}\:\mathrm{z}^{\mathrm{m}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{m}} }\:=\mathrm{2cosm}\beta. \\ $$

Question Number 138611 Answers: 2 Comments: 0

Given a function f where f(x)≥ 0for ∀x∈R. If the area U = { (x,y)∣0≤2y≤f(x), −6≤x≤−2} is u and the area V={(x,y)∣0≤y≤f(x),−2≤x≤0} is v then what the value of ∫_1 ^2 4x f(2x^2 −8) dx . (A) 5u+4v (D)2u+v (B) 4u+3v (E) u+v (C) 3u+2v

$${Given}\:{a}\:{function}\:{f}\:{where}\: \\ $$$${f}\left({x}\right)\geqslant\:\mathrm{0}{for}\:\forall{x}\in\mathbb{R}.\:{If}\:{the}\:{area} \\ $$$${U}\:=\:\left\{\:\left({x},{y}\right)\mid\mathrm{0}\leqslant\mathrm{2}{y}\leqslant{f}\left({x}\right),\:−\mathrm{6}\leqslant{x}\leqslant−\mathrm{2}\right\} \\ $$$${is}\:{u}\:{and}\:{the}\:{area}\:{V}=\left\{\left({x},{y}\right)\mid\mathrm{0}\leqslant{y}\leqslant{f}\left({x}\right),−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}\right\} \\ $$$${is}\:{v}\:{then}\:{what}\:{the}\:{value}\:{of} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\mathrm{4}{x}\:{f}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}\right)\:{dx}\:. \\ $$$$\left({A}\right)\:\mathrm{5}{u}+\mathrm{4}{v}\:\:\:\:\:\left({D}\right)\mathrm{2}{u}+{v} \\ $$$$\left({B}\right)\:\mathrm{4}{u}+\mathrm{3}{v}\:\:\:\:\left({E}\right)\:{u}+{v} \\ $$$$\left({C}\right)\:\mathrm{3}{u}+\mathrm{2}{v} \\ $$

Question Number 138608 Answers: 0 Comments: 0

Prove or disprove ∫_0 ^a x^2 e^(−x^2 ) dx=(1/e^a^2 )((a^3 /(1.3))+((2a^5 )/(1.3.5))+((2a^7 )/(1.3.5.7))+((2a^9 )/(1.3.5.7.9))+ad inf..)

$${Prove}\:{or}\:{disprove} \\ $$$$\int_{\mathrm{0}} ^{{a}} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{{e}^{{a}^{\mathrm{2}} } }\left(\frac{{a}^{\mathrm{3}} }{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}{a}^{\mathrm{5}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{2}{a}^{\mathrm{7}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{2}{a}^{\mathrm{9}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+{ad}\:{inf}..\right) \\ $$

Question Number 138631 Answers: 1 Comments: 3