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Next: 17.5 Trigonometric Substitution Up: 17. Antidifferentiation Techniques Previous: 17.3 Integration by Parts   Index

17.4 Integration by Substitution

We will now use the chain rule to find some antiderivatives. Let $g$ be a real valued function that is continuous on some interval $J$ and differentiable on the interior of $J$, and let $f$ be a function such that $f$ has an antiderivative $F$ on some interval $K$. We will suppose that $g(J)\subset K$ and $g\Big(\mbox{\rm interior}
(J)\Big)\subset\mbox{\rm interior}(K)$. It then follows that $F\circ g$ is continuous on $J$ and differentiable on $\mbox{\rm interior}(J)$, and

(F\circ g)'(t)=(F'\circ g)(t)g'(t)=(f\circ g)(t)g'(t)
\end{displaymath} (17.32)

for all $t$ in the interior of $J$; i.e., $F\circ g$ is an antiderivative for $(f\circ g)g'$ on $J$. Thus
\int f\Big( g(t)\Big)g'(t)dt=F \Big( g(t)\Big) \mbox{ where } F(u)=\int f(u)du.
\end{displaymath} (17.33)

There is a standard ritual for using (17.33) to find $\displaystyle {\int f\Big(
g(t)\Big)g'(t)dt}$ when an antiderivative $F$ can be found for $f$. We write:

Let $u=g(t)$. Then $du=g'(t)dt$ (or $\displaystyle {du={{du}\over {dt}}dt}$), so

\int f\Big( g(t)\Big)g'(t)dt=\int f(u)du=F(u)=F\Big( g(t)\Big).
\end{displaymath} (17.34)

In the first equality of (17.34) we replace $g(t)$ by $u$ and $g'(t)dt$ by $du$, and in the last step we replace $u$ by $g(t)$. Since we have never assigned any meaning to `` $du$" or `` $dt$", we should think of (17.34) just as a mnemonic device for remembering (17.33).

17.35   Example. Find $\displaystyle { \int {{\sin (\sqrt x)} \over {\sqrt x}}dx}$.

Let $u=\sqrt x$. Then $\displaystyle {du={1\over {2\sqrt x}}dx}$, so

\int {{\sin (\sqrt x)}\over {\sqrt x}}dx &=& 2\int\sin (\sqrt ...
... (u)du=-2\cos (u)\\
&=& -2\cos (\sqrt x).\mbox{ $\diamondsuit$}

Suppose we want to find $\int\sin (\sqrt x)dx$. If we had a $\sqrt x$ in the denominator, this would be a simple problem. (In fact we just considered this problem in the previous example.) We will now discuss a method of introducing the missing $\sqrt x$.

Suppose $g$ is a function on an interval $J$ such that $g'(t)$ is never zero on the interior of $J$, and suppose that $h$ is an inverse function for $g$. Then

\begin{displaymath}\Bigg(h\Big(g(x)\Big)=x\Bigg)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Bigg( h'\Big( g(x)\Big)\cdot

for all $x$ in the interior of $J$, so

\begin{displaymath}\int f\Big( g(x) \Big) dx=\int f\Big(g(x)\Big)\cdot h'\Big( g(x)\Big)\cdot

We now apply the ritual (17.34): Let $u=g(x)$. Then $du=g'(x)dx$, so

\int f\Big( g(x)\Big)dx &=& \int f\Big( g(x)\Big)h'\Big( g(x)\Big)\cdot
&=& \int f(u)h'(u)du.

If we can find an antiderivative $H$ for $fh'$, then

\begin{displaymath}\int f(u)h'(u)du=H(u)=H\Big( g(x)\Big).\end{displaymath}

We have shown that if $h$ is an inverse function for $g$, then
\int f\Big(g(x)\Big)dx=H\Big( g(x)\Big) \mbox{ where } H(u)=\int f(u)h'(u)du
\end{displaymath} (17.36)

There is a ritual associated with this result also. To find $\displaystyle {\int f\Big(
g(x)\Big) dx}$:

Let $u=g(x)$. Then $x=h(u)$ so $dx=h'(u)du$.


\int f\Big( g(x)\Big)dx=\int f(u)h'(u)du=H(u)=H\Big( g(x)\Big).
\end{displaymath} (17.37)

17.38   Example. To find $\int\sin (\sqrt x)dx$.

Let $u=\sqrt x$. Then $x=u^2$ so $dx=2u\;du$.


\begin{displaymath}\int\sin (\sqrt x)dx=\int\sin (u)\cdot 2u\;du=2\int u\sin (u)du.\end{displaymath}

We can now use integration by parts to find $\int u\sin (u)du$. Let

f(u)=u, \qquad g'(u) &=& \sin (u),\\
f'(u)=1, \qquad g(u) &=& -\cos (u).\\


\int u\sin (u)du &=& \int f(u)g'(u)du\\
&=& f(u)g(u)-\int f'(...
&=& -u\cos (u)+\int\cos (u)du\\
&=& -u\cos (u)+\sin (u).\\


\int\sin(\sqrt x)dx &=& 2\int u\sin u\; du\\
&=& -2u\cos (u)+2\sin (u)\\
&=& -2\sqrt x\cos (\sqrt x)+2\sin (\sqrt x).\\

17.39   Example. To find $\displaystyle { \int {1\over {e^x+e^{-x}}}dx}$.

Let $u=e^x$. Then $x=\ln (u)$ so $dx=\displaystyle { {1\over u}du}$.

\int {1\over {e^x+e^{-x}}}dx &=& \int {1\over {(u+{1\over u})}...
{1\over {u^2+1}}du \\
&=& \arctan (u)=\arctan (e^x).\\

17.40   Example. To find $\int t\sqrt{t+1}\;dt$.

Let $u=t+1$. Then $t=u-1$ so $dt=du$.


\int t\sqrt{t+1}\;dt &=& \int (u-1)\sqrt u\; du = \int u^{3/2}...
...2\over 3}u^{3/2}={2\over 5}(t+1)^{5/2}-{2\over

17.41   Example. To find $\displaystyle \int {{ (1-x)^{2 \over 5} \over x^{12\over 5}}dx}$.

\begin{displaymath}\int { (1-x)^{2 \over 5} \over x^{12\over 5}}dx =
\int {\left( {1-x \over x} \right)}^{2\over 5} \cdot {1\over x^2} dx. \end{displaymath}

Let $\displaystyle {u = {1-x \over x} = {1\over x} - 1}$. Then $\displaystyle {du = -{1\over x^2}dx}$, and

\int {{ (1-x)^{2 \over 5} \over x^{12\over 5}}dx} =
-\int u...
...over 5}
= -{5\over 7} \left( {1-x\over x} \right)^{7\over 5}.


\begin{displaymath}\int { (1-x)^{2 \over 5} \over x^{12\over 5}}dx =
-{5\over 7} \left( {1-x\over x} \right)^{7\over 5}.\end{displaymath}

17.42   Exercise. A Find the following antiderivatives:

$\displaystyle {\int x^2\sin (x^3)dx}$.

$\displaystyle {\int {{e^x}\over {1+e^x}}dx}$.

$\displaystyle {\int e^{\sqrt x} dx}$.

$\displaystyle { \int {{\ln (3x)}\over x}dx}$.

$\displaystyle {\int 2^x\; dx}$.

$\displaystyle {\int {{e^{2x}+e^{3x}}\over {e^{4x}}}\;dx}$.

$\displaystyle {\int x(1+\root 3 \of x )dx}$.

next up previous index
Next: 17.5 Trigonometric Substitution Up: 17. Antidifferentiation Techniques Previous: 17.3 Integration by Parts   Index
Ray Mayer 2007-09-07