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Question Number 42801 Answers: 1 Comments: 1

find f(x) = ∫_(π/4) ^(π/3) ((cosxdx)/(2cos^2 x +sin^2 x +1))

$${find}\:{f}\left({x}\right)\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{cosxdx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\:+\mathrm{1}} \\ $$

Question Number 42800 Answers: 1 Comments: 0

find ∫ ((sinx)/(1+2 cosx))dx

$${find}\:\int\:\:\:\:\:\frac{{sinx}}{\mathrm{1}+\mathrm{2}\:{cosx}}{dx} \\ $$

Question Number 42799 Answers: 0 Comments: 2

let I = ∫_0 ^(π/8) e^(−2t) cos^4 t and J =∫_0 ^(π/8) e^(−2t) sin^4 dt find the values of I andJ .

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{−\mathrm{2}{t}} \:{cos}^{\mathrm{4}} {t}\:\:\:\:{and}\:{J}\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:{sin}^{\mathrm{4}} {dt} \\ $$$${find}\:{the}\:{values}\:{of}\:{I}\:{andJ}\:. \\ $$

Question Number 42798 Answers: 1 Comments: 2

find I_n = ∫_0 ^1 x^n (√(1−x^2 ))dx

$${find}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 42797 Answers: 0 Comments: 1

let u_k = ∫_(−(π/2) +kπ) ^(−(π/2) +(k+1)π) e^(−t) cost dt 1) calculate u_k 2) let A_n =Σ_(k=0) ^n u_k find lim_(n→+∞) A_n

$${let}\:{u}_{{k}} =\:\int_{−\frac{\pi}{\mathrm{2}}\:+{k}\pi} ^{−\frac{\pi}{\mathrm{2}}\:+\left({k}+\mathrm{1}\right)\pi} \:\:{e}^{−{t}} \:{cost}\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{u}_{{k}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{u}_{{k}} \:\:\:\:\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Question Number 42796 Answers: 1 Comments: 1

calculate I = ∫_0 ^1 (x^2 /(1+x^2 )) arctan(x)dx

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

Question Number 42795 Answers: 0 Comments: 3

calculate ∫_(−∞) ^(+∞) (dx/((1+x^2 +x^4 )^2 ))

$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$

Question Number 42793 Answers: 1 Comments: 0

find f(a)= ∫_0 ^1 (dt/((a^2 +t^2 )^3 )) with a>0

$${find}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\:{with}\:{a}>\mathrm{0} \\ $$

Question Number 42792 Answers: 1 Comments: 1

find ∫_0 ^1 ((1+x^2 )/(1+x^3 ))dx

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\: \\ $$

Question Number 42791 Answers: 2 Comments: 0

find ∫ (dx/((x^(2 ) +1)(√(1+x^2 ))))

$${find}\:\:\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}\:} +\mathrm{1}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$

Question Number 42790 Answers: 1 Comments: 0

1) calculate ∫_0 ^∞ (t/(1+t^4 )) dt 2) calculate ∫_0 ^1 (t/(1+t^4 ))dt 3) calculste ∫_1 ^(+∞) (t/(1+t^4 ))dt

$$\left.\mathrm{1}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculste}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\: \\ $$

Question Number 42789 Answers: 0 Comments: 1

let u_0 =1 and u_(n+1) =u_n + (2/u_n ) study the convervence of (u_n )

$${let}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} \:={u}_{{n}} \:+\:\frac{\mathrm{2}}{{u}_{{n}} } \\ $$$${study}\:{the}\:{convervence}\:{of}\:\left({u}_{{n}} \right) \\ $$

Question Number 42788 Answers: 1 Comments: 1

calculate lim_(x→0) ((2x)/(ln(((1+x)/(1−x))))) −cosx

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

Question Number 42787 Answers: 0 Comments: 0

find lim_(x→0) (1/((sinx)^4 )){ sin((x/(1−x)))−((sinx)/(1−sinx))}

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{1}}{\left({sinx}\right)^{\mathrm{4}} }\left\{\:{sin}\left(\frac{{x}}{\mathrm{1}−{x}}\right)−\frac{{sinx}}{\mathrm{1}−{sinx}}\right\} \\ $$

Question Number 42786 Answers: 1 Comments: 1

calculate lim_(x→0) ((1−(x/(sinx)))/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} } \\ $$

Question Number 42785 Answers: 1 Comments: 1

find lim_(x→0) ((1+x −e^(arcsinx) )/x^2 )

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}+{x}\:−{e}^{{arcsinx}} }{{x}^{\mathrm{2}} } \\ $$

Question Number 42784 Answers: 0 Comments: 0

find lim_(x→0^+ ) ln(((e^(x^2 −x) −1)/x))

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

Question Number 42783 Answers: 0 Comments: 0

calculate lim_(x→(π/4)) ∣tan(2x)∣^(sin(4x))

$${calculate}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\:\:\:\mid{tan}\left(\mathrm{2}{x}\right)\mid^{{sin}\left(\mathrm{4}{x}\right)} \\ $$

Question Number 42782 Answers: 0 Comments: 0

calculate lim_(x→0^+ ) {tan((π/(2+x)))}^x

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

Question Number 42781 Answers: 1 Comments: 1

calculate lim_(x→(π/4)) ((sin(2x)sin(x−(π/4)))/(sinx −cosx))

$${calculate}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\:\frac{{sin}\left(\mathrm{2}{x}\right){sin}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{{sinx}\:−{cosx}} \\ $$

Question Number 42780 Answers: 0 Comments: 0

find lim_(x→0^+ ) (([(x+1)^2 ] −[(2x+1)^2 ])/x)

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\:\:\frac{\left[\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right]\:−\left[\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \right]}{{x}} \\ $$

Question Number 42779 Answers: 0 Comments: 1

calculate lim_(x→−∞) (x^4 +1)tan((1/x)) .

$${calculate}\:{lim}_{{x}\rightarrow−\infty} \:\:\left({x}^{\mathrm{4}} +\mathrm{1}\right){tan}\left(\frac{\mathrm{1}}{{x}}\right)\:. \\ $$

Question Number 42776 Answers: 1 Comments: 0

Solve: q^4 − 40q^2 + q + 384 = 0

$$\mathrm{Solve}:\:\:\:\:\:\mathrm{q}^{\mathrm{4}} \:−\:\mathrm{40q}^{\mathrm{2}} \:+\:\mathrm{q}\:+\:\mathrm{384}\:=\:\mathrm{0} \\ $$

Question Number 42775 Answers: 0 Comments: 2

my mother fell down and broke waist...so i ambusy for mother...age 85...i am now kolkata for mother

$${my}\:{mother}\:{fell}\:{down}\:{and}\:{broke}\:{waist}...{so}\:{i}\:{ambusy} \\ $$$${for}\:{mother}...{age}\:\mathrm{85}...{i}\:{am}\:{now}\:{kolkata}\:{for}\:{mother} \\ $$

Question Number 42773 Answers: 0 Comments: 0

let f(x) = ∫_0 ^1 (e^t /(1+x^t )) dt with 0<x<1 give f(x) at form of serie .

$${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{e}^{{t}} }{\mathrm{1}+{x}^{{t}} }\:{dt}\:\:\:\:\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${give}\:{f}\left({x}\right)\:{at}\:{form}\:{of}\:{serie}\:. \\ $$

Question Number 42772 Answers: 0 Comments: 1

mag∫2x+x^3 =

$$\mathrm{mag}\int\mathrm{2x}+\mathrm{x}^{\mathrm{3}} \\ $$$$= \\ $$

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