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

$$(F\circ g)'(t)=(F'\circ g)(t)g'(t)=(f\circ g)(t)g'(t)$$ (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

$$\int f\Big( g(t)\Big)g'(t)dt=F \Big( g(t)\Big) \mbox{ where } F(u)=\int f(u)du.$$ (17.33)

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

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

$$\int f\Big( g(t)\Big)g'(t)dt=\int f(u)du=F(u)=F\Big( g(t)\Big).$$ (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 $${ \int {{\sin (\sqrt x)} \over {\sqrt x}}dx}$$.

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

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

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

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

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

$$\int f\Big( g(x) \Big) dx=\int f\Big(g(x)\Big)\cdot h'\Big( g(x)\Big)\cdot g'(x)dx.$$

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

$$\begin{eqnarray*} \int f\Big( g(x)\Big)dx &=& \int f\Big( g(x)\Big)h'\Big( g(x)\Big)\cdot g'(x)dx\\ &=& \int f(u)h'(u)du. \end{eqnarray*}$$

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

$$\int f(u)h'(u)du=H(u)=H\Big( g(x)\Big).$$

We have shown that if $h$ is an inverse function for $g$, then

$$\int f\Big(g(x)\Big)dx=H\Big( g(x)\Big) \mbox{ where } H(u)=\int f(u)h'(u)du$$ (17.36)

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

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

Hence

$$\int f\Big( g(x)\Big)dx=\int f(u)h'(u)du=H(u)=H\Big( g(x)\Big).$$ (17.37)

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

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

Thus

$$\int\sin (\sqrt x)dx=\int\sin (u)\cdot 2u\;du=2\int u\sin (u)du.$$

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

$$\begin{eqnarray*} f(u)=u, \qquad g'(u) &=& \sin (u),\\ f'(u)=1, \qquad g(u) &=& -\cos (u).\\ \end{eqnarray*}$$

Then

$$\begin{eqnarray*} \int u\sin (u)du &=& \int f(u)g'(u)du\\ &=& f(u)g(u)-\int f'(... ...\\ &=& -u\cos (u)+\int\cos (u)du\\ &=& -u\cos (u)+\sin (u).\\ \end{eqnarray*}$$

Hence

$$\begin{eqnarray*} \int\sin(\sqrt x)dx &=& 2\int u\sin u\; du\\ &=& -2u\cos (u)+2\sin (u)\\ &=& -2\sqrt x\cos (\sqrt x)+2\sin (\sqrt x).\\ \end{eqnarray*}$$

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

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

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

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

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

Hence

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

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

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

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

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

Thus

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

17.42   Exercise. A Find the following antiderivatives:

a)

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

b)

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

c)

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

d)

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

e)

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

f)

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

g)

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