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Question Number 59193 Answers: 1 Comments: 0

Question Number 59190 Answers: 0 Comments: 0

find ∫_0 ^1 (x^2 /(1−cosx))dx .

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{cosx}}{dx}\:. \\ $$

Question Number 59188 Answers: 2 Comments: 4

let f(x)=x−(√(4−x^2 )) and g(x) =((2 +(√(x−3)))/(2−(√(x−3)))) 1) find D_f ,D_g and D_(fog) and determine fog(x) 2) calculate gof(x) and give D_(gof) 3) calculate ∫_(−(1/2)) ^(1/2) f(x)dx 4) calculate ∫_4 ^5 g(x)dx .

$${let}\:{f}\left({x}\right)={x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:{and}\:{g}\left({x}\right)\:=\frac{\mathrm{2}\:+\sqrt{{x}−\mathrm{3}}}{\mathrm{2}−\sqrt{{x}−\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:\:{D}_{{f}} \:\:,{D}_{{g}} \:\:\:{and}\:{D}_{{fog}} \:\:\:\:\:{and}\:\:{determine}\:{fog}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{gof}\left({x}\right)\:{and}\:{give}\:{D}_{{gof}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:\:\:\: \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{\mathrm{5}} \:{g}\left({x}\right){dx}\:. \\ $$

Question Number 59187 Answers: 0 Comments: 2

calculate lim_(x→0) ((ln(arctan(1+x))−ln((π/4)))/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left({arctan}\left(\mathrm{1}+{x}\right)\right)−{ln}\left(\frac{\pi}{\mathrm{4}}\right)}{{x}^{\mathrm{2}} } \\ $$

Question Number 59186 Answers: 0 Comments: 2

calculate lim_(x→0) ((arctan{ln(1+x)})/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{arctan}\left\{{ln}\left(\mathrm{1}+{x}\right)\right\}}{{x}^{\mathrm{2}} } \\ $$

Question Number 59185 Answers: 0 Comments: 1

calculate ∫_0 ^1 (x/(sinx))dx let f(x) =sinx ⇒f(x) =f(0) +xf^′ (0) +(x^2 /2)f^((2)) (0)+(x^3 /(3!))f^((3)) (0) +o(x^4 ) but f(0) =0 f^′ (x) =cosx ⇒f^′ (0)=1 f^((2)) (x) =−sinx ⇒f^((2)) (0)=0 f^((3)) (0) =−cos(x) ⇒f^((3)) (0) =−1 ⇒sinx =x−(x^3 /6) +o(x^4 ) ⇒ x−(x^3 /3) ≤sinx ≤x for x ∈]0,1] (1/x) ≤(1/(sinx)) ≤(1/(x−(x^3 /6))) ⇒1≤(x/(sinx)) ≤(1/(1−(x^2 /6))) ⇒ 1 ≤ ∫_0 ^1 (x/(sinx)) dx ≤ ∫_0 ^1 (dx/(1−(x^2 /6))) ∫_0 ^1 (dx/(1−(x^2 /6))) =_(x =(√6)t) ∫_0 ^(1/(√6)) (((√6)dt)/(1−t^2 )) =((√6)/2) ∫_0 ^(1/(√6)) ((1/(1−t)) +(1/(1+t)))dt =((√6)/2)[ln∣((1+t)/(1−t))∣]_0 ^(1/(√6)) =((√6)/2) { ln(((1+(1/(√6)))/(1−(1/(√6)))))} =((√6)/2) ln((((√6)+1)/((√6)−1))) ⇒ 1≤ ∫_0 ^1 (x/(sinx)) dx ≤ ((√6)/2)ln((((√6)+1)/((√6)−1))) .

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{sinx}}{dx} \\ $$$${let}\:{f}\left({x}\right)\:={sinx}\:\Rightarrow{f}\left({x}\right)\:={f}\left(\mathrm{0}\right)\:+{xf}^{'} \left(\mathrm{0}\right)\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:+{o}\left({x}^{\mathrm{4}} \right) \\ $$$${but}\:{f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\:\:{f}^{'} \left({x}\right)\:={cosx}\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)=\mathrm{1}\:\:\:\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−{sinx}\:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−{cos}\left({x}\right)\:\Rightarrow{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−\mathrm{1}\:\Rightarrow{sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$$\left.{x}\left.−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\leqslant{sinx}\:\leqslant{x}\:\:\:\:{for}\:{x}\:\in\right]\mathrm{0},\mathrm{1}\right]\:\:\:\frac{\mathrm{1}}{{x}}\:\leqslant\frac{\mathrm{1}}{{sinx}}\:\leqslant\frac{\mathrm{1}}{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}\:\Rightarrow\mathrm{1}\leqslant\frac{{x}}{{sinx}}\:\leqslant\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:\Rightarrow \\ $$$$\mathrm{1}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{{sinx}}\:{dx}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:=_{{x}\:=\sqrt{\mathrm{6}}{t}} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}} \:\:\:\:\frac{\sqrt{\mathrm{6}}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}} \:\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}} \:\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\left\{\:{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}}{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}}\right)\right\}\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:{ln}\left(\frac{\sqrt{\mathrm{6}}+\mathrm{1}}{\sqrt{\mathrm{6}}−\mathrm{1}}\right)\:\Rightarrow \\ $$$$\mathrm{1}\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{sinx}}\:{dx}\:\leqslant\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{6}}+\mathrm{1}}{\sqrt{\mathrm{6}}−\mathrm{1}}\right)\:. \\ $$$$ \\ $$

Question Number 59184 Answers: 0 Comments: 0

find ∫ ((cos(3x))/(ch(2x)))dx .

$${find}\:\:\:\int\:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{{ch}\left(\mathrm{2}{x}\right)}{dx}\:. \\ $$

Question Number 59183 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((arctan(1+ix))/(2+x^2 )) dx

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

Question Number 59182 Answers: 0 Comments: 2

let f(x) =arctan(1+ix) determine Re(f(x)) and Im(f(x))dx

$${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{1}+{ix}\right) \\ $$$${determine}\:{Re}\left({f}\left({x}\right)\right)\:{and}\:{Im}\left({f}\left({x}\right)\right){dx} \\ $$

Question Number 59181 Answers: 1 Comments: 1

Question Number 59177 Answers: 0 Comments: 1

∫ ((cos 2x)/(cos x)) dx =

$$\int\:\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}}\:{dx}\:= \\ $$

Question Number 59175 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ ((1−cos(x))/x^2 )dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}\: \\ $$

Question Number 59174 Answers: 0 Comments: 0

calculate ∫∫_([0,2]^2 ) (x+1−(√y))(y+1−(√x))dxdy

$${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{2}\right]^{\mathrm{2}} } \:\:\:\:\left({x}+\mathrm{1}−\sqrt{{y}}\right)\left({y}+\mathrm{1}−\sqrt{{x}}\right){dxdy}\: \\ $$

Question Number 59172 Answers: 0 Comments: 1

calculate ∫∫_([1,3]^2 ) (x+y)ln(x^2 +y^2 )dxdy

$${calculate}\:\int\int_{\left[\mathrm{1},\mathrm{3}\right]^{\mathrm{2}} } \:\:\:\:\left({x}+{y}\right){ln}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){dxdy}\: \\ $$

Question Number 59171 Answers: 0 Comments: 0

let f(x,y) ((arctan(x+2y))/(x +y^2 )) calculate (∂f/∂x)(x,y) , (∂f/∂y)(x,y),(∂^2 f/∂x^2 )(x,y), (∂^2 f/∂y^2 )(x,y) , (∂^2 f/(∂x∂y))(x,y) (∂^2 f/(∂y∂x))(x,y)

$${let}\:{f}\left({x},{y}\right)\:\:\frac{{arctan}\left({x}+\mathrm{2}{y}\right)}{{x}\:+{y}^{\mathrm{2}} } \\ $$$${calculate}\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:\:,\:\frac{\partial{f}}{\partial{y}}\left({x},{y}\right),\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right),\:\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }\left({x},{y}\right)\:,\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}\left({x},{y}\right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}}\left({x},{y}\right) \\ $$

Question Number 59170 Answers: 1 Comments: 0

Iff(x)= determinant (((sec x),(cos x),(sec^2 x+cosec x cot x)),((cos^2 x),(cos^2 x),( cosec^2 x)),(( 1),(cos^2 x),( cos^2 x))) then ∫_( 0) ^(π/2) f(x) dx =

$$\mathrm{If}{f}\left({x}\right)=\begin{vmatrix}{\mathrm{sec}\:{x}}&{\mathrm{cos}\:{x}}&{\mathrm{sec}^{\mathrm{2}} {x}+\mathrm{cosec}\:{x}\:\mathrm{cot}\:{x}}\\{\mathrm{cos}^{\mathrm{2}} {x}}&{\mathrm{cos}^{\mathrm{2}} {x}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{cosec}^{\mathrm{2}} {x}}\\{\:\:\:\mathrm{1}}&{\mathrm{cos}^{\mathrm{2}} {x}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{cos}^{\mathrm{2}} {x}}\end{vmatrix} \\ $$$$\mathrm{then}\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:{f}\left({x}\right)\:{dx}\:= \\ $$

Question Number 59169 Answers: 0 Comments: 1

calculate A_n =∫∫_([(1/n),n[^2 ) e^(−x^2 −3y^2 ) (√(x^2 +3y^2 ))dxdy and find lim_(n→+∞) A_n

$${calculate}\:{A}_{{n}} =\int\int_{\left[\frac{\mathrm{1}}{{n}},{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\:{e}^{−{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} } \sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}} }{dxdy} \\ $$$${and}\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Question Number 59168 Answers: 1 Comments: 0

calculate ∫_(−1) ^1 ((x^2 +3)/((√(1+x)) +(√(1−x)))) dx

$${calculate}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$

Question Number 59167 Answers: 0 Comments: 0

find the value of Σ_(n=1) ^∞ ((n^2 +1)/(n^3 (n+1)^2 ))

$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}^{\mathrm{2}} \:+\mathrm{1}}{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 59166 Answers: 0 Comments: 0

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