Integration by Substitution
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Integration by substitution (often called "usubstitution") is essentially the reverse of the chain rule just like integration itself is the inverse function of differentiation.
For example, consider this problem:
Find the derivative of
#f(x)# :#f(x)=sqrt(3x^2+2x5)# The chain rule says that we begin with the "outside function" which is:
#f(u)=u^(1/2)# where#u=3x^2+2x5# #f'(u)=(1/2)u^(1/2)# , but we now must multiply by#u'=(6x+2)#
giving us the final answer (after substituting for#u# ):#f'(x)=(6x+2)/(2sqrt(3x^2+2x5)# Now consider the integration problem, find the indefinite integral:
#int(6x+2)/(2sqrt(3x^2+2x5))dx# We recognize this as the exact reverse of the previous problem.
We are being asked to integrate:
#int(du)/(2sqrt(u))# =#int(1/2)u^(1/2)du# where
#u = 3x^2+2x5# and#du=(6x+2)dx# which is already present in the problem.Clearly, the solution is
#f(x)=sqrt(3x^2+2x5)# .The key to making this work, and something I have glossed over here by selecting a simplified example where the exact value of
#du# was already present, is that it only works if we have#du# in the integral.For example, it is not possible to use this method on:
#int1/(sqrt(3x^2+2x5))dx# because the differential,
#du=3x^2+2x5# ,does not appear in the problem. This is probably the most common mistake students make when trying to apply integration by
#u# substitution. 
#sin(e^x)+c# , where c is a constantExplanation,
Integration by Substitution,
#inte^x*cos(e^x)dx# let's assume
#e^x=t# then,
#e^x*dx=dt# #intcos(t)dt# #sin(t)+c# , where c is a constantSubstituting
#t=e^x# , we finally get#sin(e^x)+c# , where c is a constant 
This expression can be integrated using u = substitution where
#u = x^3+1# and#du = 3x^2dx^# .We take the
#int(x^2sqrt(x^3+1))dx# and let#u = (x^3+1)# .
This makes#du=3x^2dx# . We now see that we are missing a constant of 3 on the#x^2# term. We can multiply the term by 3 provided that we multiply the entire expression by#1/3# so that the expression balances. We now see that the#sqrt# is enclosing the term#x^3+1# . A#sqrt# is equivalent to the power#1/2# . We take the antiderivative this exponent which is#3/2# giving us the final answer#1/3(x^3+1)^(3/2)# .