illustration of integration techniques

The following integral is an example that illustrates many integration techniques.

Problem. Determine the antiderivative of $\sqrt{\tan x}$.

. We start with substitution (http://planetmath.org/IntegrationBySubstitution):

$u$$=\sqrt{\tan x}$

$u^2$$=\tan x$

$2udu$$={\sec }^{2}xdx$

Using the Pythagorean identity ${\tan }^{2}x+1={\sec }^{2}x$, we obtain:

$2udu$$=({\tan }^{2}x+1)dx$

$2udu$$=(u^4+1)dx$

${\displaystyle \frac{2u}{u^4+1}}du$$=dx$

Thus,

${\displaystyle \int \sqrt{\tan x}𝑑x}$$={\displaystyle \int u\frac{2u}{u^4+1}𝑑u}$

$={\displaystyle \int \frac{2u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)}𝑑u}.$

For this last integral, we use the method of partial fractions (http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod):

${\displaystyle \frac{2u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)}}$$={\displaystyle \frac{A+Bu}{u^2-u\sqrt{2}+1}}+{\displaystyle \frac{C+Du}{u^2+u\sqrt{2}+1}}$

$2u^2$$=(A+Bu)(u^2+u\sqrt{2}+1)+(C+Du)(u^2-u\sqrt{2}+1)$

$=(B+D)u^3+(A+C+(B-D)\sqrt{2})u^2+(B+D+(A-C)\sqrt{2})u+A+C$

From this, we obtain the following system of equations:

This can be into two smaller systems of equations:

It is clear that the first system yields $A=C=0$, and it can easily be verified that $B=\frac{1}{\sqrt{2}}$ and $D=\frac{-1}{\sqrt{2}}$. Therefore,

${\displaystyle \int \sqrt{\tan x}𝑑x}$$={\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{u^2-u\sqrt{2}+1}𝑑u}-{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{u^2+u\sqrt{2}+1}𝑑u}$

$={\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{u^2-u\sqrt{2}+\frac{1}{2}+\frac{1}{2}}𝑑u}-{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{u^2+u\sqrt{2}+\frac{1}{2}+\frac{1}{2}}𝑑u}$

$={\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{{(u-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}𝑑u}-{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{u}{{(u+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}𝑑u}.$

Now we make the following substitutions:

Note that we have $v+\frac{1}{\sqrt{2}}=u=w-\frac{1}{\sqrt{2}}$. Therefore,

${\displaystyle \int \sqrt{\tan x}𝑑x}$$={\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{v+\frac{1}{\sqrt{2}}}{v^2+\frac{1}{2}}𝑑v}-{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{w-\frac{1}{\sqrt{2}}}{w^2+\frac{1}{2}}𝑑w}$

$={\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{v}{v^2+\frac{1}{2}}𝑑v}-{\displaystyle \frac{1}{2}}{\displaystyle \int \frac{dv}{v^2+\frac{1}{2}}}-{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \int \frac{w}{w^2+\frac{1}{2}}𝑑w}+{\displaystyle \frac{1}{2}}{\displaystyle \int \frac{dw}{w^2+\frac{1}{2}}}.$

For the first and third integrals in the last expression, note that the numerator is a of the derivative of the denominator. For these, we use the formula

$$\int \frac{k{f}^{\prime }(x)}{f(x)}𝑑x=k\ln |f(x)|.$$

For the second and fourth integrals in the last expression, we use the formula

$$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan \left(\frac{x}{a}\right)$$

with $a=\frac{1}{\sqrt{2}}$. Hence,

${\displaystyle \int \sqrt{\tan x}𝑑x}$$={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left(v^2+{\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}\arctan (v\sqrt{2})-{\displaystyle \frac{1}{2\sqrt{2}}}\ln \left(w^2+{\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}\arctan (w\sqrt{2})+K$

$={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{v^2+\frac{1}{2}}{w^2+\frac{1}{2}}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}(\arctan (v\sqrt{2})+\arctan (w\sqrt{2}))+K$

$={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{{(u-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}{{(u+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}\left(\arctan \left[\left(u-{\displaystyle \frac{1}{\sqrt{2}}}\right)\sqrt{2}\right]+\arctan \left[\left(u+{\displaystyle \frac{1}{\sqrt{2}}}\right)\sqrt{2}\right]\right)+K$

$={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{u^2-u\sqrt{2}+1}{u^2+u\sqrt{2}+1}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}[\arctan (u\sqrt{2}-1)+\arctan (u\sqrt{2}+1)]+K$

$={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}}\right)+{\displaystyle \frac{1}{\sqrt{2}}}[\arctan (\sqrt{2\tan x}-1)+\arctan (\sqrt{2\tan x}+1)]+K.$

(We use $K$ for the constant of integration to avoid confusion with $C$ from the system of equations.)