Use Table of Integrals to evaluate the integral. x3sin⁡(x2+10)dx.

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Cheryl King · Accepted Answer

∫x3sin⁡(x2+10)dx Let x2+10=t 2xdx=dt 12∫(t−10)sin⁡tdt =12[∫t⋅sin⁡tdt−10∫sin⁡tdt] =12[(sin⁡t−tcos⁡t+c)−10(−cos⁡t)] =12[sin⁡t−tcos⁡t+10cos⁡t]+C =12[sin⁡(x2+10)+10cos⁡(x2+10)−(x2+10)cos⁡(x2+10)]+C

hysgubwyri3 · Accepted Answer

∫x3sin⁡(x2+10)&amp;nbsp;dx&amp;nbsp;&amp;nbsp;Let&#039;s place the expression 2 * x under the differential&#039;s sign, i.e.2x&amp;nbsp;dx&amp;nbsp;=d(x2),t=x2&amp;nbsp;The initial integral can then be expressed as follows: The&amp;nbsp;∫t⋅sin⁡(t+10)2&amp;nbsp;dt&amp;nbsp;&amp;nbsp;∫x⋅sin⁡(x+10)2&amp;nbsp;dx&amp;nbsp;&amp;nbsp;formula for integration by parts:&amp;nbsp;∫U(x)⋅dV(x)=U(x)⋅V(x)−∫V(x)⋅dU(x)&amp;nbsp;Put&amp;nbsp;U=x&amp;nbsp;dV=sin⁡(x+10)2⋅&amp;nbsp;dx&amp;nbsp;&amp;nbsp;Then:&amp;nbsp;dU=dx&amp;nbsp;V=−cos⁡(x+10)2&amp;nbsp;Thus:&amp;nbsp;∫x⋅sin⁡(x+10)2&amp;nbsp;dx&amp;nbsp;=−x⋅cos⁡(x+10)2−∫cos⁡(x+10)2&amp;nbsp;dx&amp;nbsp;=−x⋅cos⁡(x+10)2+∫cos⁡(x+10)2&amp;nbsp;dx&amp;nbsp;&amp;nbsp;Find the integral&amp;nbsp;∫cos⁡(x+10)2&amp;nbsp;dx&amp;nbsp;=sin⁡(x+10)2&amp;nbsp;Result:&amp;nbsp;x⋅sin⁡(x+10)2=−x⋅cos⁡(x+10)2+sin⁡(x+10)2+C&amp;nbsp;To write down the final answer, it remains to substitute&amp;nbsp;x2&amp;nbsp;instead of t.&amp;nbsp;−x2⋅cos⁡(x2+10)2+sin⁡(x2+10)2+C

Vasquez · Accepted Answer

∫x3×sin⁡(x2+10)dx∫t×sin⁡(t)−10sin⁡(t)2dt12×∫t×sin⁡(t)−10sin⁡(t)dt12×(∫t×sin⁡(t)dt−∫10sin⁡(t)dt)12×(−t×cos⁡(t)+sin⁡(t)+10cos⁡(t))12×(−(x2+10)×cos⁡(x2+10)+sin⁡(x2+10)+10cos⁡(x2+10))cos⁡(x2+10)×(−x2−10)+sin⁡(x2+10)2+5cos⁡(x2+10)Solution:cos⁡(x2+10)×(−x2−10)+sin⁡(x2+10)2+5cos⁡(x2+10)+C

Use Table of Integrals to evaluate the integral. x^{3} \sin(x^{2}

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