How do you find the integral #int cos(2x+1)dx# using substitution?

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Answer 1 · 2018-04-14T03:57:07

#intcos(2x+1)dx=1/2sin(2x+1)+C#

#u=2x+1#

#(du)/dx=2#

#du=2dx#

Solve for #dx,# yielding

#1/2du=dx#

Then, using the substitution and factoring the constant out, we have

#1/2intcosudu=1/2sinu+C#

Rewriting in terms of #x# yields

#intcos(2x+1)dx=1/2sin(2x+1)+C#