U-substitution

Introduction

The method of $u$-substitution is used to simplify the function you are integrating so that you can easily recognize it's antiderivative. This method is closely related to the chain rule for derivatives.

One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master $u$-substitution is to work out as many problems as possible. This will help you: (1) understand the $u$-substitution method and (2) correctly identify the necessary substitution.

NOTE: After you plug-in your substitution, all of the $x$'s in your integral should be gone. The only variables remaining in your integral should be $u$'s.

Warm-Up

Evaluate the following indefinite integrals.

1)   $\int (8x+5)e^{4x^{2}+5x+3}~dx$

Solution:
Let $u=4x^{2}+5x+3$. Then, $du=(8x+5)~dx$.
Plugging these into our integral, we get $\int e^{u}~du$, which we know how to integrate.
So, we get
${\begin{array}{rcl}\displaystyle {\int (8x+5)e^{4x^{2}+5x+3}~dx}&=&\displaystyle {\int e^{u}~du}\\&&\\&=&\displaystyle {e^{u}+C}\\&&\\&=&\displaystyle {e^{4x^{2}+5x+3}+C}.\\\end{array}}$

Final Answer:
$e^{4x^{2}+5x+3}+C$

2)   $\int {\frac {x}{\sqrt {1-2x^{2}}}}~dx$

Solution:
Let $u=1-2x^{2}$. Then, $du=-4x~dx$. Hence, ${\frac {du}{-4}}=x~dx$.
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int {\frac {x}{\sqrt {1-2x^{2}}}}~dx}&=&\displaystyle {\int -{\frac {1}{4}}~u^{-{\frac {1}{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{2}}u^{\frac {1}{2}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C}.\\\end{array}}$

Final Answer:
$-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C$

3)   $\int {\frac {\sin(\ln x)}{x}}~dx$

Solution:

Final Answer:
$-\cos(\ln x)+C$

4)   $\int xe^{x^{2}}~dx$

Solution:
Let $u=x^{2}$. Then, $du=2x~dx$ and ${\frac {du}{2}}=x~dx$.
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int xe^{x^{2}}~dx}&=&\displaystyle {\int {\frac {1}{2}}e^{u}~du}\\&&\\&=&\displaystyle {{\frac {1}{2}}e^{u}+C}\\&&\\&=&\displaystyle {{\frac {1}{2}}e^{x^{2}}+C}.\\\end{array}}$

Final Answer:
${\frac {1}{2}}e^{x^{2}}+C$