intlimits_0^{{{left( {frac{pi }{2}} right)}^{ | vedclass

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(D) Let $I = \int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $. Substitute $t = x^3$,then $dt = 3x^2 dx

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$1/3$

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(D) Let $I = \int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $. Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} dt$. When $x = 0$,$t = 0$. When $x = (\frac{\pi}{2})^{1/3}$,$t = \frac{\pi}{2}$. Substituting these into the integral: $I = \int\limits_0^{\frac{\pi}{2}} {t \cdot \sin(t) \cdot \frac{1}{3} dt} = \frac{1}{3} \int\limits_0^{\frac{\pi}{2}} {t \sin(t) dt}$. Using integration by parts $\int u dv = uv - \int v du$,let $u = t$ and $dv = \sin(t) dt$. Then $du = dt$ and $v = -\cos(t)$. $I = \frac{1}{3} [ -t \cos(t) |_0^{\frac{\pi}{2}} - \int\limits_0^{\frac{\pi}{2}} {-\cos(t) dt} ]$. $I = \frac{1}{3} [ (- \frac{\pi}{2} \cos(\frac{\pi}{2}) + 0) + \int\limits_0^{\frac{\pi}{2}} \cos(t) dt ]$. Since $\cos(\frac{\pi}{2}) = 0$,the first term is $0$. $I = \frac{1}{3} [ \sin(t) |_0^{\frac{\pi}{2}} ] = \frac{1}{3} [ \sin(\frac{\pi}{2}) - \sin(0) ] = \frac{1}{3} [ 1 - 0 ] = \frac{1}{3}$.

$\int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $ $=$

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