int_0^a {frac{{{x^4},dx}}{{{{({a^2} + {x^2})} | vedclass

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Question

(A) Let $I = \int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} $. Put $x = a 	an 	heta $,then $dx = a \sec^2 	heta \, d	heta $. When $x = 0$,

Vedclass · Accepted Answer

$\frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$

Vedclass · Answer

(A) Let $I = \int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} $. Put $x = a 	an 	heta $,then $dx = a \sec^2 	heta \, d	heta $. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{4}$. $I = \int_0^{\pi/4} \frac{a^4 	an^4 	heta \cdot a \sec^2 	heta \, d	heta}{(a^2 + a^2 	an^2 	heta)^4} = \int_0^{\pi/4} \frac{a^5 	an^4 	heta \sec^2 	heta \, d	heta}{a^8 \sec^8 	heta} = \frac{1}{a^3} \int_0^{\pi/4} \sin^4 	heta \cos^2 	heta \, d	heta$. Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we have $\sin^4 	heta \cos^2 	heta = (\frac{1 - \cos 2	heta}{2})^2 (\frac{1 + \cos 2	heta}{2}) = \frac{1}{8} (1 - 2\cos 2	heta + \cos^2 2	heta)(1 + \cos 2	heta) = \frac{1}{8} (1 - \cos 2	heta - \cos^2 2	heta + \cos^3 2	heta)$. Using $\cos^2 2	heta = \frac{1 + \cos 4	heta}{2}$ and $\cos^3 2	heta = \frac{3\cos 2	heta + \cos 6	heta}{4}$,$I = \frac{1}{8a^3} \int_0^{\pi/4} (1 - \cos 2	heta - \frac{1 + \cos 4	heta}{2} + \frac{3\cos 2	heta + \cos 6	heta}{4}) \, d	heta = \frac{1}{32a^3} \int_0^{\pi/4} (4 - 4\cos 2	heta - 2 - 2\cos 4	heta + 3\cos 2	heta + \cos 6	heta) \, d	heta = \frac{1}{32a^3} \int_0^{\pi/4} (2 - \cos 2	heta - 2\cos 4	heta + \cos 6	heta) \, d	heta$. $I = \frac{1}{32a^3} [2	heta - \frac{\sin 2	heta}{2} - \frac{\sin 4	heta}{2} + \frac{\sin 6	heta}{6}]_0^{\pi/4} = \frac{1}{32a^3} [2(\frac{\pi}{4}) - \frac{1}{2}(1) - \frac{1}{2}(0) + \frac{1}{6}(-1)] = \frac{1}{32a^3} [\frac{\pi}{2} - \frac{1}{2} - \frac{1}{6}] = \frac{1}{32a^3} [\frac{\pi}{2} - \frac{2}{3}] = \frac{1}{16a^3} [\frac{\pi}{4} - \frac{1}{3}]$.

$\int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} = $

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