Question #49555

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Answer 1 · 2017-11-18T04:49:55

Let $x^2 = theta$

$intx^3cos(x^2)dx = int theta^(3/2)costheta (d theta)/(2theta^(1/2))$

$=int thetacostheta d theta$

$int udv = uv- int v du$

$u = theta, dv = cos theta d theta$ then $v = sin theta$

$(1/2)int thetacostheta d theta = (1/2)(theta sintheta - int sintheta d theta)$

$=(1/2)(theta sin theta +cos theta+C)$
$=(1/2)(x^2 sin(x^2) + cos(x^2) + C)$

Let us differentiate and check the result

If $f = x^2 sin(x^2) + cos(x^2) + C$

$(df)/(dx) = \cancel(2xsin(x^2)) + x^2 cos(x^2) (2x)-\cancel(2xsin(x^2))$

$= 2x^3cos(x^2)$

$1/2 (df)/(dx) = x^3cos(x^2)$

Hence the solution above is correct.