Page:Calculus Made Easy.pdf/249

Do you remember meeting ${\dfrac {dx}{\sqrt {1-x^{2}}}}$? it is got by differentiating $y=\arcsin x$ (see p. 171); hence its integral is $\arcsin x$, and so

$\int {\sqrt {1-x^{2}}}\,dx={\frac {x{\sqrt {1-x^{2}}}}{2}}+{\tfrac {1}{2}}\arcsin x+C$.

You can try now some exercises by yourself; you will find some at the end of this chapter.

Substitution. This is the same dodge as explained in Chap. IX., p. 67. Let us illustrate its application to integration by a few examples.

(1) $\int {\sqrt {3+x}}\,dx$.
Let

$3+x=u,\quad dx=du$;

replace

$\int u^{\frac {1}{2}}\,du={\tfrac {2}{3}}u^{\frac {3}{2}}={\tfrac {2}{3}}(3+x)^{\frac {3}{2}}$.

(2) Let $\int {\dfrac {dx}{\epsilon ^{x}+\epsilon ^{-x}}}$;
so that

$\int {\frac {dx}{\epsilon ^{x}+\epsilon ^{-x}}}=\int {\frac {du}{\epsilon ^{x}(\epsilon ^{x}+\epsilon ^{-x})}}=\int {\frac {du}{u\left(u+{\dfrac {1}{u}}\right)}}=\int {\frac {du}{u^{2}+1}}$.

${\dfrac {du}{1+u^{2}}}$ is the result of differentiating $\arctan x$.

Hence the integral is $\arctan \epsilon ^{x}$.

(3) $\int {\dfrac {dx}{x^{2}+2x+3}}=\int {\dfrac {dx}{x^{2}+2x+1+2}}=\int {\dfrac {dx}{(x+1)^{2}+({\sqrt {2}})^{2}}}$.