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Question Number 53118 Answers: 1 Comments: 0

If a<∫_0 ^(2π) (1/(10+3 cos x)) dx<b, then the ordered pair (a, b) is

$$\mathrm{If}\:{a}<\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{10}+\mathrm{3}\:\mathrm{cos}\:{x}}\:{dx}<{b},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{ordered}\:\mathrm{pair}\:\left({a},\:{b}\right)\:\mathrm{is} \\ $$

Question Number 53114 Answers: 0 Comments: 1

let A_n =∫_0 ^∞ ((x sin(nx))/((x^2 +n^2 )^2 ))dx with n integr natural not 0 1) find the value of A_n 2) study the convergence of Σ A_n

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}\:{sin}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{with}\:{n}\:{integr}\:{natural}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{A}_{{n}} \\ $$

Question Number 53113 Answers: 0 Comments: 1

let I =∫_(−∞) ^(+∞) ((t+1)/((t^2 −t+1)^2 ))dt find value of I .

$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}+\mathrm{1}}{\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${find}\:{value}\:{of}\:{I}\:. \\ $$

Question Number 53112 Answers: 1 Comments: 0

calculate ∫_0 ^π ((1+2sinx)/(3 +2cosx))dx let A =∫_0 ^π ((1+2sinx)/(3 +2cosx))dx changement tan((x/2))=t give A =∫_0 ^∞ ((1+((4t)/(1+t^2 )))/(3+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫_0 ^∞ ((1+t^2 +4t)/((1+t^2 )^2 (((3+3t^2 +2−2t^2 )/(1+t^2 )))))dt =2 ∫_0 ^∞ ((t^2 +4t +1)/((1+t^2 )(5+t^2 )))dt let decompose F(t)=((t^2 +4t+1)/((t^2 +1)(t^2 +5))) F(t)=((at +b)/(t^2 +1)) +((ct +d)/(t^2 +5)) ⇒(at+b)(t^2 +5)+(ct+d)(t^2 +1) =t^2 +4t +1 ⇒ at^3 +5at +bt^2 +5b +ct^3 +ct +dt^2 +d =t^2 +4t +1 ⇒ (a+c)t^3 +(b+d)t^2 +(5a+c)t +5b +d =t^2 +4t +1 ⇒a+c=0 and b+d=1 and 5a+c =4 and 5b+d =1 ⇒c=−a ⇒a=1 ⇒c=−1 we have d=1−b ⇒5b +1−b =1 ⇒b=0 ⇒d=1 ⇒ F(t)=(t/(t^2 +1)) +((−t +1)/(t^2 +5)) ⇒ A =2 ∫_0 ^∞ F(t)dt =∫_0 ^∞ ((2t)/(t^2 +1))dt +∫_0 ^∞ ((−2t +2)/(t^2 +5))dt =[ln(((t^2 +1)/(t^2 +5)))]_0 ^(+∞) +2 ∫_0 ^∞ (dt/(t^2 +5)) =ln(5) + 2 ∫_0 ^∞ (dt/(t^2 +5)) but ∫_0 ^∞ (dt/(t^2 +5))dt =_(t =(√5)u ) ∫_0 ^∞ (((√5)du)/(5(1+u^2 ))) =(1/(√5)) [artanu]_0 ^(+∞) =(π/(2(√5))) ⇒ A =ln(5) +(π/(2(√5))) .

Question Number 53108 Answers: 1 Comments: 0

((x + 2))^(1/3) − ((x − 3))^(1/3) > (1/2)

$$\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:+\:\mathrm{2}}\:\:−\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:−\:\mathrm{3}}\:\:\:>\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 53086 Answers: 0 Comments: 1

JEE Main chemistry errors by Resonance. Today is the last day for raising objectiions. You need to tap once to select and tap the text again to open hyperlink.

$$\mathrm{JEE}\:\mathrm{Main}\:\mathrm{chemistry}\:\mathrm{errors}\:\mathrm{by} \\ $$$$\mathrm{Resonance}.\:\mathrm{Today}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{day} \\ $$$$\mathrm{for}\:\mathrm{raising}\:\mathrm{objectiions}. \\ $$$$\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{tap}\:\mathrm{once}\:\mathrm{to}\:\mathrm{select}\:\mathrm{and}\:\mathrm{tap} \\ $$$$\mathrm{the}\:\mathrm{text}\:\mathrm{again}\:\mathrm{to}\:\mathrm{open}\:\mathrm{hyperlink}. \\ $$

Question Number 53081 Answers: 3 Comments: 0