Question and Answers Forum

IntegrationQuestion and Answers: Page 128

Question Number 123967 Answers: 1 Comments: 0

Question Number 123937 Answers: 2 Comments: 0

...nice calculus... prove that:: ∫_((−π)/4) ^(π/4) (((π−4x)tan(x))/(1−tan(x)))dx=^(???) πln(2)−(π^2 /4)

$$\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:\:\:\:{calculus}... \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\pi−\mathrm{4}{x}\right){tan}\left({x}\right)}{\mathrm{1}−{tan}\left({x}\right)}{dx}\overset{???} {=}\pi{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$

Question Number 123921 Answers: 3 Comments: 0

∫ (1/( (√x) (x+1)((tan^(−1) (√x))^2 +9)))dx

$$\int\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:\left({x}+\mathrm{1}\right)\left(\left(\mathrm{tan}^{−\mathrm{1}} \sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{9}\right)}{dx} \\ $$

Question Number 123900 Answers: 2 Comments: 0

... advanced integral... prove that :: Ω=∫_0 ^( ∞) ((sin^2 (x))/(x^2 (1+x^2 )))dx =(π/4)(1+(π/e^2 ))

$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:\:{integral}... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+\frac{\pi}{{e}^{\mathrm{2}} }\right) \\ $$

Question Number 123896 Answers: 1 Comments: 0

∫_0 ^π ((sin^2 x)/( (√x))) dx

$$\:\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\:\sqrt{{x}}}\:{dx} \\ $$

Question Number 123920 Answers: 2 Comments: 0

∫ (((x−1)(√(x^4 +2x^3 −x^2 +2x+1)))/(x^2 (x+1))) dx ?

$$\:\int\:\frac{\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:{dx}\:? \\ $$

Question Number 123866 Answers: 2 Comments: 0

To mr mjs sir. integral lover ∫_0 ^(π/4) ((tan ((π/4)−x))/(cos^2 x (√(tan^3 x+tan^2 x+tan x)))) dx =?

$$\:{To}\:{mr}\:{mjs}\:{sir}.\: \\ $$$${integral}\:{lover}\: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)}{\mathrm{cos}\:^{\mathrm{2}} {x}\:\sqrt{\mathrm{tan}\:^{\mathrm{3}} {x}+\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:{x}}}\:{dx}\:=?\: \\ $$

Question Number 123859 Answers: 1 Comments: 0

∫sect e^t dt

$$\int{sect}\:{e}^{{t}} \:{dt} \\ $$

Question Number 123793 Answers: 2 Comments: 0

∫ ((4x−1)/(2x^2 −3x+2)) dx ?

$$\:\:\int\:\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\:{dx}\:? \\ $$

Question Number 123769 Answers: 0 Comments: 1

Question Number 123768 Answers: 0 Comments: 0

Question Number 123767 Answers: 1 Comments: 0

Question Number 123764 Answers: 1 Comments: 0

.... advanced calculus ... prove that:::: Σ_(n=1) ^∞ {((ζ(2n+1))/(4^(n ) (2n+1)))}=ln(2)−γ γ:: euler−mascheroni constant

$$\:\:\:\:\:\:\:\:\:\:\:\:....\:{advanced}\:\:{calculus}\:... \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}:::: \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\zeta\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}^{{n}\:} \:\left(\mathrm{2}{n}+\mathrm{1}\right)}\right\}={ln}\left(\mathrm{2}\right)−\gamma \\ $$$$\:\:\:\:\:\:\:\:\gamma::\:\:{euler}−{mascheroni} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{constant} \\ $$

Question Number 123745 Answers: 0 Comments: 0

...advanced calculus... evaluation of: Ω=∫_0 ^( (π/2)) sin(x)log(sin(x))dx by using the euler beta and gamma function: β(p,(1/2))=2∫_0 ^(π/2) sin^(2p−1) (x)dx ((dβ(p,(1/2)))/dp) =2∫_0 ^(π/2) 2sin^(2p−1) (x)ln(sin(x))dx =4∫_0 ^(π/2) sin^(2p−1) (x)ln(sin(x))dx Ω=(1/4)[((dβ(p,(1/2)))/dp)]_(p=1) ((d(β(p,(1/2))))/dp)=(√π)[((Γ′(p)Γ(p+(1/2))−Γ′(p+(1/2))Γ(p))/(Γ^2 (p+(1/2))))]_(p=1) Ω=(1/4)((√π)[((Γ^′ (1)Γ((3/2))−Γ′((3/2))Γ(1))/(Γ^2 ((3/2))))]) =((√π)/4)[((−γ(((√π)/2))−((√π)/2)(2−γ−2ln(2)))/(π/4))] =((√π)/4)∗((√π)/2)(((−γ−2+γ+2ln(2))/(π/4))) =ln(2)−1 ... m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:\:{calculus}... \\ $$$$\:\:{evaluation}\:\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}\left({x}\right){log}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:{by}\:{using}\:{the}\:{euler}\:{beta}\:{and}\:{gamma}\:{function}: \\ $$$$\:\:\:\:\:\:\:\:\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){dx} \\ $$$$\:\:\:\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\right]_{{p}=\mathrm{1}} \\ $$$$\:\frac{{d}\left(\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}{{dp}}=\sqrt{\pi}\left[\frac{\Gamma'\left({p}\right)\Gamma\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Gamma'\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({p}\right)}{\Gamma^{\mathrm{2}} \left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right]_{{p}=\mathrm{1}} \\ $$$$\:\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\pi}\left[\frac{\Gamma^{'} \left(\mathrm{1}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\right]\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\left[\frac{−\gamma\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)−\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{2}−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)}{\frac{\pi}{\mathrm{4}}}\right] \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\ast\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{−\gamma−\mathrm{2}+\gamma+\mathrm{2}{ln}\left(\mathrm{2}\right)}{\frac{\pi}{\mathrm{4}}}\right) \\ $$$$={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:...\:{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\: \\ $$

Question Number 123710 Answers: 2 Comments: 0

calculate ∫_1 ^(√3) (dx/((x^2 +1)^2 (x+2)^5 ))

$${calculate}\:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} } \\ $$

Question Number 123703 Answers: 1 Comments: 0