intlimits_0^infty {frac{{{x^3}}}{{1 + x + 2{ | vedclass

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(D) Let $I = \int_0^\infty  {\frac{{{x^3}}}{{1 + x + 2{x^2} + 2{x^3} + {x^4} + {x^5}}}} dx$. Factor the denominator: $1 + x + 2x^2 + 2x^3 + x^4 + x^5

Vedclass · Accepted Answer

$\frac{{\pi  - 1}}{4}$

Vedclass · Answer

(D) Let $I = \int_0^\infty  {\frac{{{x^3}}}{{1 + x + 2{x^2} + 2{x^3} + {x^4} + {x^5}}}} dx$. Factor the denominator: $1 + x + 2x^2 + 2x^3 + x^4 + x^5 = (1+x)(1+x^2+x^4) = (1+x)(1+x^2)^2$. So,$I = \int_0^\infty  {\frac{{{x^3}}}{{(1 + x){{(1 + {x^2})}^2}}}} dx$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. As $x 	o 0, 	heta 	o 0$ and as $x 	o \infty, 	heta 	o \frac{\pi}{2}$. $I = \int_0^{\frac{\pi}{2}} \frac{	an^3 	heta \sec^2 	heta}{(1 + 	an 	heta) \sec^4 	heta} d	heta = \int_0^{\frac{\pi}{2}} \frac{	an^3 	heta}{(1 + 	an 	heta) \sec^2 	heta} d	heta = \int_0^{\frac{\pi}{2}} \frac{\sin^3 	heta}{\cos 	heta + \sin 	heta} d	heta$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 	heta}{\cos 	heta + \sin 	heta} d	heta$. Adding the two expressions for $I$: $2I = \int_0^{\frac{\pi}{2}} \frac{\sin^3 	heta + \cos^3 	heta}{\sin 	heta + \cos 	heta} d	heta = \int_0^{\frac{\pi}{2}} (\sin^2 	heta + \cos^2 	heta - \sin 	heta \cos 	heta) d	heta$. $2I = \int_0^{\frac{\pi}{2}} (1 - \frac{1}{2} \sin 2	heta) d	heta = [	heta + \frac{1}{4} \cos 2	heta]_0^{\frac{\pi}{2}} = (\frac{\pi}{2} + \frac{1}{4} \cos \pi) - (0 + \frac{1}{4} \cos 0) = \frac{\pi}{2} - \frac{1}{4} - \frac{1}{4} = \frac{\pi}{2} - \frac{1}{2} = \frac{\pi - 1}{2}$. Thus,$I = \frac{\pi - 1}{4}$.

$\int\limits_0^\infty {\frac{{{x^3}}}{{1 + x + 2{x^2} + 2{x^3} + {x^4} + {x^5}}}} dx$

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