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Question Number 143899 Answers: 3 Comments: 0

Question Number 143898 Answers: 0 Comments: 0

Question Number 143892 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((cos^4 (2x))/((x^4 −x^2 +1)^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}^{\mathrm{4}} \left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{4}} \:−\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 143890 Answers: 2 Comments: 0

if a+b+c=π tanA+tanB+tanC =tanA.tanB.tanC FAILED TO CALCULATE

$${if}\:{a}+{b}+{c}=\pi\:\:\: \\ $$$$\:{tanA}+{tanB}+{tanC}\:={tanA}.{tanB}.{tanC} \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$

Question Number 143897 Answers: 1 Comments: 0

Question Number 143878 Answers: 3 Comments: 0

y′′′=2xy′′

$$\mathrm{y}'''=\mathrm{2xy}'' \\ $$

Question Number 143875 Answers: 0 Comments: 1

Question Number 143874 Answers: 0 Comments: 0

y′′−x×y′′′+y′′′^3 =0

$$\mathrm{y}''−\mathrm{x}×\mathrm{y}'''+\mathrm{y}'''^{\mathrm{3}} =\mathrm{0} \\ $$

Question Number 143863 Answers: 0 Comments: 1

Question Number 143860 Answers: 1 Comments: 0

calculate ∫_0 ^∞ x e^(−x^2 ) log(1+e^x )dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$

Question Number 143858 Answers: 1 Comments: 0

if A≥0, B≥0, A+B=(Π/3) then find minimum and maximum of tan A.tan B

$$\mathrm{if}\:\mathrm{A}\geqslant\mathrm{0},\:\mathrm{B}\geqslant\mathrm{0},\:\mathrm{A}+\mathrm{B}=\frac{\Pi}{\mathrm{3}}\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\: \\ $$$$\mathrm{of}\:\mathrm{tan}\:\mathrm{A}.\mathrm{tan}\:\mathrm{B}\: \\ $$

Question Number 143857 Answers: 0 Comments: 0

if cos^4 θsec^2 α, (1/2), sin^4 θcosec^2 α are in A.P. then cos^8 θsec^6 α, (1/2), sin^8 θcosec^6 α are in which progression?

$$\mathrm{if}\:\mathrm{cos}^{\mathrm{4}} \theta\mathrm{sec}\:^{\mathrm{2}} \alpha,\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{sin}^{\mathrm{4}} \theta\mathrm{cosec}^{\mathrm{2}} \alpha\: \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{A}.\mathrm{P}.\:\mathrm{then} \\ $$$$\:\mathrm{cos}^{\mathrm{8}} \theta\mathrm{sec}^{\mathrm{6}} \alpha,\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{sin}^{\mathrm{8}} \theta\mathrm{cosec}^{\mathrm{6}} \alpha \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{which}\:\mathrm{progression}? \\ $$

Question Number 143868 Answers: 2 Comments: 1

Question Number 143850 Answers: 0 Comments: 0

Question Number 143846 Answers: 1 Comments: 0

find the sum (((0!)^2 )/(1!)) +(((1!)^2 )/(3!)) +(((2!)^2 )/(5!)) +.....

$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\:\frac{\left(\mathrm{0}!\right)^{\mathrm{2}} }{\mathrm{1}!}\:+\frac{\left(\mathrm{1}!\right)^{\mathrm{2}} }{\mathrm{3}!}\:+\frac{\left(\mathrm{2}!\right)^{\mathrm{2}} }{\mathrm{5}!}\:+..... \\ $$

Question Number 143832 Answers: 3 Comments: 0

x^2 + y^2 = (((56)/(13)))^2 and x + ((5y)/(12)) = ((56)/(12)) find x+y=?

$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{56}}{\mathrm{13}}\right)^{\mathrm{2}} \:{and}\:\:{x}\:+\:\frac{\mathrm{5}{y}}{\mathrm{12}}\:=\:\frac{\mathrm{56}}{\mathrm{12}} \\ $$$${find}\:\:{x}+{y}=? \\ $$

Question Number 143856 Answers: 0 Comments: 2

if ((cos^4 x)/(cos^2 y))+((sin^4 x)/(sin^2 y))=1then find ((cos^4 y)/(cos^2 x))+((sin^4 y)/(sin^2 x))

$$\mathrm{if}\:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}^{\mathrm{2}} \mathrm{y}}+\frac{\mathrm{sin}^{\mathrm{4}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{y}}=\mathrm{1then}\:\mathrm{find}\: \\ $$$$\:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{y}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{sin}^{\mathrm{4}} \mathrm{y}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$

Question Number 143822 Answers: 0 Comments: 0

x^4 +ax^2 +bx+c=0 let x=(t/s) ⇒ t^4 +as^2 t^2 +bs^3 t+cs^4 =0 let t=p+h ⇒ p^4 +4hp^3 +6h^2 p^2 +4h^3 p+h^4 +as^2 (p^2 +2hp+h^2 ) +bs^3 (p+h)+cs^4 =0 ⇒ p^4 +4hp^3 +(6h^2 +as^2 )p^2 +(4h^3 +2ahs^2 +bs^3 )p +(h^4 +as^2 h^2 +bs^3 h+cs^4 )=0 let ((√2)p^2 +Ap+B)^2 =(p^2 +mp+k)^2 ⇒ 2(√2)A−2m=4h 2(√2)B+A^2 −m^2 −2k = 6h^2 +as^2 2AB−2mk=4h^3 +2ah+bs^3 B^2 −k^2 =h^4 +as^2 h^2 +bs^3 h+cs^4 let m=0 ⇒ A=(√2)h 2(√2)B=4h^2 +as^2 +2k 4h^3 +ahs^2 +2hk=4h^3 +2ah+bs^3 ⇒ (as^2 +2k−2a)h=bs^3 (4h^3 +as^2 +2k)^2 −8k^2 =8(h^4 +as^2 h^2 +bs^3 h+cs^4 ) let k=a ⇒ ah=bs ⇒ (((4b^2 h)/a^2 )+a+((2a^3 h^2 )/b^2 ))^2 −((8b^4 h^4 )/a^2 ) =8((b^4 /a^4 )+((2b^2 )/a)+c) ⇒ ((16b^4 h^2 )/a^4 )+((4a^6 h^4 )/b^4 ) +((8b^2 h)/a)+32ah^3 +((4a^4 h^2 )/b^2 ) −((8b^4 h^4 )/a^2 )=λ ⇒ (((4a^6 )/b^4 )−((8b^4 )/a^2 ))h^4 +32ah^3 +(((16b^4 )/a^4 )+((4a^4 )/b^2 ))h^2 +((8b^2 h)/a)−λ=0 ...

$$\:{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{t}}{{s}} \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} +{as}^{\mathrm{2}} {t}^{\mathrm{2}} +{bs}^{\mathrm{3}} {t}+{cs}^{\mathrm{4}} =\mathrm{0} \\ $$$${let}\:{t}={p}+{h}\:\:\Rightarrow \\ $$$${p}^{\mathrm{4}} +\mathrm{4}{hp}^{\mathrm{3}} +\mathrm{6}{h}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{3}} {p}+{h}^{\mathrm{4}} \\ $$$$+{as}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{2}{hp}+{h}^{\mathrm{2}} \right) \\ $$$$+{bs}^{\mathrm{3}} \left({p}+{h}\right)+{cs}^{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{4}} +\mathrm{4}{hp}^{\mathrm{3}} +\left(\mathrm{6}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} \right){p}^{\mathrm{2}} \\ $$$$+\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ahs}^{\mathrm{2}} +{bs}^{\mathrm{3}} \right){p} \\ $$$$+\left({h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${let}\:\:\left(\sqrt{\mathrm{2}}{p}^{\mathrm{2}} +{Ap}+{B}\right)^{\mathrm{2}} =\left({p}^{\mathrm{2}} +{mp}+{k}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}\sqrt{\mathrm{2}}{A}−\mathrm{2}{m}=\mathrm{4}{h} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{B}+{A}^{\mathrm{2}} −{m}^{\mathrm{2}} −\mathrm{2}{k} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{6}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} \\ $$$$\mathrm{2}{AB}−\mathrm{2}{mk}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ah}+{bs}^{\mathrm{3}} \\ $$$${B}^{\mathrm{2}} −{k}^{\mathrm{2}} ={h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \\ $$$${let}\:\:{m}=\mathrm{0}\:\:\Rightarrow \\ $$$${A}=\sqrt{\mathrm{2}}{h} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{B}=\mathrm{4}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} +\mathrm{2}{k} \\ $$$$\mathrm{4}{h}^{\mathrm{3}} +{ahs}^{\mathrm{2}} +\mathrm{2}{hk}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ah}+{bs}^{\mathrm{3}} \\ $$$$\Rightarrow\:\left({as}^{\mathrm{2}} +\mathrm{2}{k}−\mathrm{2}{a}\right){h}={bs}^{\mathrm{3}} \\ $$$$\left(\mathrm{4}{h}^{\mathrm{3}} +{as}^{\mathrm{2}} +\mathrm{2}{k}\right)^{\mathrm{2}} \\ $$$$\:\:−\mathrm{8}{k}^{\mathrm{2}} =\mathrm{8}\left({h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \right) \\ $$$${let}\:\:{k}={a} \\ $$$$\Rightarrow\:\:{ah}={bs} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{4}{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} }+{a}+\frac{\mathrm{2}{a}^{\mathrm{3}} {h}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\frac{\mathrm{8}{b}^{\mathrm{4}} {h}^{\mathrm{4}} }{{a}^{\mathrm{2}} } \\ $$$$\:\:=\mathrm{8}\left(\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}+{c}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{16}{b}^{\mathrm{4}} {h}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{4}{a}^{\mathrm{6}} {h}^{\mathrm{4}} }{{b}^{\mathrm{4}} } \\ $$$$+\frac{\mathrm{8}{b}^{\mathrm{2}} {h}}{{a}}+\mathrm{32}{ah}^{\mathrm{3}} +\frac{\mathrm{4}{a}^{\mathrm{4}} {h}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\:\:−\frac{\mathrm{8}{b}^{\mathrm{4}} {h}^{\mathrm{4}} }{{a}^{\mathrm{2}} }=\lambda \\ $$$$\Rightarrow\:\left(\frac{\mathrm{4}{a}^{\mathrm{6}} }{{b}^{\mathrm{4}} }−\frac{\mathrm{8}{b}^{\mathrm{4}} }{{a}^{\mathrm{2}} }\right){h}^{\mathrm{4}} +\mathrm{32}{ah}^{\mathrm{3}} \\ $$$$\:\:\:+\left(\frac{\mathrm{16}{b}^{\mathrm{4}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{4}{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }\right){h}^{\mathrm{2}} +\frac{\mathrm{8}{b}^{\mathrm{2}} {h}}{{a}}−\lambda=\mathrm{0} \\ $$$$... \\ $$

Question Number 144084 Answers: 1 Comments: 0

if a+(1/a)=−1 a^(1234567891011) +(1/a^(1110987654321) )=??

$${if}\:\:{a}+\frac{\mathrm{1}}{{a}}=−\mathrm{1} \\ $$$$\:{a}^{\mathrm{1234567891011}} +\frac{\mathrm{1}}{{a}^{\mathrm{1110987654321}} }=?? \\ $$$$ \\ $$

Question Number 144086 Answers: 1 Comments: 0

If sin t + cos t = (2/3) then cosec t+sec t =?

$$\mathrm{If}\:\mathrm{sin}\:\mathrm{t}\:+\:\mathrm{cos}\:\:\mathrm{t}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{cosec}\:\mathrm{t}+\mathrm{sec}\:\mathrm{t}\:=? \\ $$

Question Number 144085 Answers: 1 Comments: 1