How do you use Integration by Substitution to find #intx*sin(x^2)dx#?

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Answer 1 · 2014-09-26T09:19:31

Begin by letting making a u-substitution.

Let #u=x^2#

#du=2x dx#

#(du)/2=x dx =># This can be substituted in the original integral

#intx*sin(x^2)dx=intsin(u)*(du)/2=1/2intsin(u)du#

#=1/2[-cos(u)]+C#

#=-1/2[cos(u)]+C#

Rewrite in terms of #x#

#=-1/2[cos(x^2)]+C#