### Integration substitutions

The *chain rule* can also be ‘run backward’, and is called **change of variables** or **substitution** or sometimes **u-substitution**. Some examples of what happens are straightforward,
but others are less obvious. It is at this point that the capacity to
*recognize derivatives* from past experience becomes very
helpful.

*Example* Since (by the chain rule)
$${d\over dx} e^{\sin x}=\cos x\;e^{\sin x}$$
then we can anticipate that
$$\int \cos x\;e^{\sin x}\; dx= e^{\sin x}+C$$

*Example* Since (by the chain rule)
$${d\over dx} \sqrt{x^5+3x}={1\over 2}(x^5+3x)^{-1/2}\cdot(5x^4+3)$$
then we can anticipate that
$$\int {1\over 2}(5x^4+3)(x^5+3x)^{-1/2}\;dx=\sqrt{x^5+3x}+C$$

Very often it happens that things are *off by a
constant*. This should not deter a person from recognizing the
possibilities. For example: since, by the chain rule,
$${d\over dx}\sqrt{5+e^x}={1\over 2}(5+e^x)^{-1/2}\cdot e^x$$
then
$$\int e^x\,(5+e^x)^{-1/2}\;dx=2\int {1\over 2}e^x(5+e^x)^{-1/2}\;dx=
2\,\sqrt{5+e^x}+C$$
Notice how for ‘bookkeeping purposes’ we put the ${1\over 2}$ into
the integral (to make the constants right there) and put a
compensating $2$ outside.

*Example*: Since (by the chain rule)
$${d\over dx}\sin^7(3x+1)=7\cdot \sin^6(3x+1)\cdot \cos (3x+1)\cdot
3$$
then we have
\begin{multline*}\int \cos(3x+1)\,\sin^6(3x+1)\;dx\\
={1\over 21}\int 7\cdot 3\cdot \cos(3x+1) \sin^6(3x+1)\; dx=
{1\over 21} \sin^7(3x+1)+C
\end{multline*}

#### Exercises

- $\int \cos x\,\sin x\,dx=?$
- $\int 2x \;e^{x^2}\;dx=?$
- $\int 6x^5\;e^{x^6}\; dx=?$
- $\int { \cos x \over \sin x }\; dx=?$
- $\int \cos x e^{\sin x}\; dx=?$
- $\int { 1 \over 2\sqrt{x }}e^{\sqrt{x}}\; dx=?$
- $\int \cos x \sin^5 x\; dx=?$
- $\int \sec^2 x \tan^7 x\; dx=?$
- $\int (3\cos x+ x)\,e^{6\sin x+x^2}\,dx=?$
- $\int e^x\sqrt{e^x+1}\,dx=?$

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##### Calculus Refresher

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