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.