Integration by Substitution

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U-substitution
5:11 — by Khan Academy

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  • Integration by substitution (often called "u-substitution") 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+2x-5)#

    The chain rule says that we begin with the "outside function" which is:

    #f(u)=u^(1/2)# where #u=3x^2+2x-5#

    #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+2x-5)#

    Now consider the integration problem, find the indefinite integral:

    #int(6x+2)/(2sqrt(3x^2+2x-5))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+2x-5# and #du=(6x+2)dx# which is already present in the problem.

    Clearly, the solution is #f(x)=sqrt(3x^2+2x-5)#.

    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+2x-5))dx#

    because the differential,

    #du=3x^2+2x-5#,

    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 constant

    Explanation,

    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 constant

    Substituting #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)#.

Questions

  • Eddie answered · 3 weeks ago