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17.6 Substitution in Integrals

Let $f$ be a nice function on an interval $[a,b]$. Then if $F$ is any antiderivative for $f$, we have

\begin{displaymath}\int_a^b f=F\mid_a^b =F(b)-F(a),\end{displaymath}

by the fundamental theorem of calculus. We saw in (17.32) that under suitable hypotheses on $g$, $F\circ g$ is an antiderivative for $(f\circ g)g'$. Hence

\begin{displaymath}\int_a^b f\Big(g(t)\Big)g'(t)dt=F\circ
g\mid_a^b=F\Big\vert _{g(a)}^{g(b)}=\int_{g(a)}^{g(b)}f(u)du.\end{displaymath}

Hence we can find $\displaystyle {\int_a^b f\Big( g(t)\Big)g'(t)dt}$ by the following ritual:

Let $u=g(t)$. When $t=a$ then $u=g(a)$ and when $t=b$ then $u=g(b)$. Also $du=g'(t)dt$. Hence

\begin{displaymath}\int_a^b f\Big( g(t)\Big) g'(t)dt=\int_{g(a)}^{g(b)}
f(u)du=F(u)\Big\vert _{g(a)}^{g(b)}.\end{displaymath}

17.51   Example. To find $\displaystyle {\int_{\pi^2}^{4\pi^2}{{\sin(\sqrt x)}\over {\sqrt x}}dx}$.

Let $u=\sqrt x$. When $x=\pi^2$, then $u=\pi$, and when $x=4\pi^2$, then $u=2\pi$. Also $\displaystyle {du={1\over {2\sqrt x}}dx}$, so

\begin{eqnarray*}
\int_{\pi^2}^{4\pi^2}{{\sin (\sqrt x)}\over {\sqrt x}}dx &=& 2...
...pi^{2\pi} \\
&=& -2\Big( \cos(2\pi)-\cos (\pi)\Big)=-2(1+1)=-4.
\end{eqnarray*}



We saw in (17.36) that if $h$ is an inverse function for $g$, then an antiderivative for $f\circ g$ is $H\circ g$, where $H$ is an antiderivative for $f\cdot h'$. Thus

\begin{displaymath}\int_a^b f\Big( g(t)\Big)dt =H\circ g\mid_a^b=H\Big\vert _{g(a)}^{g(b)}.\end{displaymath}

The ritual for finding $\displaystyle { \int_a^b f\Big(g(t)\Big)dt}$ in this case is:

Let $u=g(t)$. Then $t=h(u)$ and $dt=h'(u)du$. When $t=a$ then $u=g(a)$, and when $t=b$ then $u=g(b)$. Thus

\begin{displaymath}\int_a^b f\Big( g(t)\Big)dt
=\int_{g(a)}^{g(b)}f(u)h'(u)du=H(u)\Big\vert _{g(a)}^{g(b)}\end{displaymath}

where $H$ is an antiderivative for $fh'$.

17.52   Example. To find $\displaystyle {\int_0^{\ln(\sqrt 3)}{1\over {e^x+e^{-x}}}dx}$.

Let $u=e^x$. When $x=0$ then $u=1$, and when $x=\ln(\sqrt 3)$ then $u=\sqrt 3$. Also $x=\ln (u)$, so $\displaystyle { dx={1\over u}du}$.

\begin{eqnarray*}
\int_0^{\ln(\sqrt 3)}{1\over {e^x+e^{-x}}}dx &=& \int_1^{\sqrt...
...)-\arctan (1) \\
&=& {\pi\over 3}-{\pi\over 4}={\pi\over {12}}.
\end{eqnarray*}



17.53   Exercise. A Find the following integrals:

a)
$\displaystyle {\int_0^1 x^2(x^3+1)^3 dx}$.

b)
$\displaystyle {\int_0^{3/2} {1\over {\sqrt{9-x^2}}}dx}$.

c)
$\displaystyle {\int_0^1 x\sqrt{1-x}\;dx}$.

17.54   Exercise. A Find the area of the shaded region, bounded by the ellipse $\displaystyle {
{{x^2}\over
4}+y^2=1}$ and the lines $x=\pm 1$.
\psfig{file=ch17b.eps,width=2.5in}

17.55   Example. In practice I would find many of the antiderivatives and integrals discussed in this chapter by computer. For example, using Maple, I would find

> int(sqrt(a^2+x^2),x);

\begin{displaymath}
{\displaystyle \frac {1}{2}} {x} \sqrt {{a}^{2} + {x}^{2}}...
...ln} \left( \!  {x}
+ \sqrt {{a}^{2} + {x}^{2}}  \! \right)
\end{displaymath}

> int(sin(sqrt(x)),x=0..Pi^2);

\begin{displaymath}
2 { \pi}
\end{displaymath}

> int(sqrt(4 - x^2),x=-1..1);

\begin{displaymath}
\sqrt {3} + {\displaystyle \frac {2}{3}} { \pi}
\end{displaymath}

> int( (sec(x))^3,x);

\begin{displaymath}
{\displaystyle \frac {1}{2}} {\displaystyle \frac {{\rm sin...
...1}{2}} 
{\rm ln}( {\rm sec}( {x} ) + {\rm tan}( {x} ) )
\end{displaymath}

> int(exp(a*x)*cos(b*x),x);

\begin{displaymath}
{\displaystyle \frac {{a} {\rm e}^{( {a} {x} )} {\rm co...
...( {a} {x} )} {\rm sin}( {b} {x} )}{{a}^{2} + {b}
^{2}}}
\end{displaymath}


next up previous index
Next: 17.7 Rational Functions Up: 17. Antidifferentiation Techniques Previous: 17.5 Trigonometric Substitution   Index
Ray Mayer 2007-09-07