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Ω= ∫_0 ^( ∞) (( x − sin (x ))/x^( 3) )dx −−− solution−−− Ω=^(I.B.P) [ ((−1)/(2 x^( 2) )) (x−sin(x))]_0 ^∞ +(1/2) ∫_0 ^( ∞) ((1−cos (x))/x^( 2) )dx = (1/2) ∫_0 ^( ∞) (( 2sin^( 2) ((x/2)))/x^( 2) )dx=∫_0 ^( ∞) ((sin^( 2) ((x/2)))/x^( 2) )dx =^((x/2) = α) (1/2)∫_0 ^( ∞) ((sin^( 2) ( α))/α^( 2) ) dα = (1/2) [((−1)/α) sin^( 2) (α)]_0 ^∞ +(1/2)∫_0 ^( ∞) ((sin(2α))/α)dα =^(2α=ϕ) (1/2) ∫_0 ^( ∞) (( sin(ϕ ))/ϕ) dϕ =(π/4) −− Ω= (π/4) −−−

$$ \\ $$$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}\:−\:{sin}\:\left({x}\:\right)}{{x}^{\:\mathrm{3}} }{dx} \\ $$$$−−−\:{solution}−−− \\ $$$$\:\:\:\:\:\Omega\overset{\mathscr{I}.\mathscr{B}.\mathscr{P}} {=}\:\left[\:\frac{−\mathrm{1}}{\mathrm{2}\:{x}^{\:\mathrm{2}} }\:\left({x}−{sin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{cos}\:\left({x}\right)}{{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\:\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\overset{\frac{{x}}{\mathrm{2}}\:=\:\alpha} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{2}} \left(\:\alpha\right)}{\alpha^{\:\mathrm{2}} }\:{d}\alpha\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{−\mathrm{1}}{\alpha}\:{sin}^{\:\mathrm{2}} \left(\alpha\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}\alpha\right)}{\alpha}{d}\alpha \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{2}\alpha=\varphi} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left(\varphi\:\right)}{\varphi}\:{d}\varphi\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−\:\:\:\:\:\Omega=\:\frac{\pi}{\mathrm{4}}\:\:−−− \\ $$$$ \\ $$$$ \\ $$

Question Number 163024 Answers: 0 Comments: 2

calcul en fonction de n Σ_(k=0) ^(2n) (_k ^(2n) )^(n−2k)

$${calcul}\:{en}\:{fonction}\:{de}\:{n} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(_{{k}} ^{\mathrm{2}{n}} \right)^{{n}−\mathrm{2}{k}} \\ $$

Question Number 163020 Answers: 0 Comments: 2

cos^(2n) (x)+sin^(2n) (x)=((4^n +1)/5^n ) prove that

$$\boldsymbol{\mathrm{cos}}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{sin}}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{4}^{\boldsymbol{\mathrm{n}}} +\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$

Question Number 163014 Answers: 2 Comments: 0

given (a/b)=(c/d), find an expression for (a/(a+b))

$${given}\:\frac{{a}}{{b}}=\frac{{c}}{{d}},\:{find}\:{an}\:{expression} \\ $$$${for}\:\:\frac{{a}}{{a}+{b}} \\ $$

Question Number 163028 Answers: 1 Comments: 0

Question Number 163008 Answers: 2 Comments: 0

Calculate ∫ (((2−4sin x cos x)(1+sin 2x))/(sin^4 2x+64 cos^4 2x)) dx

$$\:{Calculate}\: \\ $$$$\:\:\:\int\:\frac{\left(\mathrm{2}−\mathrm{4sin}\:{x}\:\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{2}{x}+\mathrm{64}\:\mathrm{cos}\:^{\mathrm{4}} \mathrm{2}{x}}\:{dx}\: \\ $$$$ \\ $$

Question Number 163002 Answers: 0 Comments: 0

prove that i:Σ_(n=0) ^∞ (((−1 )^( n) )/((n +(1/2))cosh(n+(1/2))π)) =(π/4) ii: ∫_0 ^( 1) (( sin( π x ))/(x^( x) ( 1−x )^( 1−x) )) (dx/(1+x)) =(π/4) −−−

$$ \\ $$$$\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:{i}:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\:\right)^{\:{n}} }{\left({n}\:+\frac{\mathrm{1}}{\mathrm{2}}\right){cosh}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:{ii}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{sin}\left(\:\pi\:{x}\:\right)}{{x}^{\:{x}} \left(\:\mathrm{1}−{x}\:\right)^{\:\mathrm{1}−{x}} }\:\frac{{dx}}{\mathrm{1}+{x}}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:−−− \\ $$

Question Number 163000 Answers: 1 Comments: 0