F. $\int \frac {1}{\sqrt {x^2 - a^2}} dx = ln |x + \sqrt {x^2 - a^2}| + C$ Proof
Put $x =a sec \theta$ then $dx= a sec \theta tan \theta d\theta $
Therefore
$\int \frac {1}{\sqrt {x^2 -a^2}} dx $
$=\int sec \theta d\theta $
$=ln |sec \theta + tan \theta| + C $
Now
$sec \theta = \frac {x}{a} $
$tan \theta = \sqrt {sec^2 \theta -1 } = \sqrt {\frac {x^2}{a^2} -1} $
Therefore
$=ln | \frac {x}{a} + \sqrt {\frac {x^2}{a^2} -1}| + C_1 = ln |x + \sqrt {x^2 - a^2}| - ln |a| +C_1 = ln |x + \sqrt {a^2 + x^2}| + C$
The above formulas are very useful to solve the various types of Integration Problems
Problem Type I
$\int \frac {1}{ax^2 + bx + c} dx$
This integral can be converted to the form A,B, C given above and can be evaluated using the formula
Example 1
$\int \frac {1}{x^2 -6x + 25} dx$ Solution
$\int \frac {1}{x^2 -6x + 25} dx= \int \frac {1}{(x -3)^2 + 4^2} dx$
Substituting y=x-3, or dy=dx, we get
$=\int \frac {1}{y^2 + 4^2} dy$
Now this can be easily evaluated using the formula A with a =4
$\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$
Hence
$=\frac {1}{4} \tan ^{-1} (\frac {y}{4}) + C$
Substituting back the value of y
$=\frac {1}{4} \tan ^{-1} (\frac {x-3}{4}) + C$
Problem Type II
$\int \frac {1}{\sqrt {ax^2 + bx + c}} dx$
This integral can be converted to the form D,E,F given above and can be evaluated using the formula
Problem Type III
$\int \frac {px + q}{ax^2 + bx + c} dx$
Here we can decompose this integral using the below formula
$px+q = A \frac {d}{dx} (ax^2 + bx + c) + B$
Then integral is converted as
$=A \int \frac {\frac {d}{dx} (ax^2 + bx + c)} {ax^2 + bx + c} dx + B \int \frac {1}{ax^2 + bx + c} dx $
The first integral is of the form
$\int \frac {f'(x)}{f(x)} dx$ which can be easily evaluated
The second integral is of the form of problem type I
$\int \frac {1}{ax^2 + bx + c} dx $ which can be easily evaluated
Problem Type IV
$\int \frac {px + q}{\sqrt{ax^2 + bx + c}} dx$
Here we can decompose this integral using the below formula
$px+q = A \frac {d}{dx} (ax^2 + bx + c) + B$
Then integral is converted as
$=A \int \frac {\frac {d}{dx} (ax^2 + bx + c)} {ax^2 + bx + c} dx + B \int \frac {1}{\sqrt {ax^2 + bx + c}} dx $
The first integral is of the form
$\int \frac {f'(x)}{f(x)^{1/2}} dx$ which can be easily evaluated
The second integral is of the form of problem type II
$\int \frac {1}{\sqrt {ax^2 + bx + c}} dx$ which can be easily evaluated
