Integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x)

[Entertainment Unit]

Homework Statement

If $a \neq 0$, evaluate the integral

$$\int \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$

(Hint: Make the substitution $u = \tan x$ and consider separately the cases where
$b^2 - 4ac$ is positive, zero, or negative.)

The Attempt at a Solution

$$\int \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$

$$\int \frac {\sec^2~x} {\sec^2~x} \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$

$$\int \frac {\sec^2~x~dx} {a~\tan^2~x + b~\tan~x + c}$$

Let $u = \tan~x \Rightarrow dx = \frac {du} {\sec^2~x}$

$$\int \frac {sec^2~x} {au^2 + bu + c} \frac {du} {\sec^2~x}$$

$$\begin{equation} \tag{1} \int \frac {du} {au^2 + bu + c} \end{equation}$$

$$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \frac c a}$$

$$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \left( \frac b {2a} \right)^2 - \left(\frac b {2a} \right)^2 + \frac c a}$$

$$\frac 1 a \int \frac {du} { \left (u + \frac b {2a} \right)^2 - \frac {b^2 - 4ac} {(2a)^2}}$$

$$\frac 1 a \int \frac {du} { \left(u + \frac b {2a} \right)^2 - \left (\frac {\sqrt {b^2 - 4ac}} {2a} \right)^2}$$

$$\frac 1 a \frac 1 {\frac {2 \sqrt {b^2 - 4ac} } {2a}} \ln \left| \frac {u + \frac b {2a} - \frac {\sqrt {b^2 - 4ac}} {2a}} {u + \frac b {2a} + \frac {\sqrt {b^2 - 4ac}} {2a}} \right| + C$$

$$\frac 1 {\sqrt {b^2 - 4ac}} \ln \left| \frac {u + \frac {b - \sqrt {b^2 - 4ac}} {2a}} {u + \frac {b + \sqrt {b^2 - 4ac}} {2a}} \right| + C$$

$$\frac 1 {\sqrt {b^2 - 4ac}} \ln \left| \frac {\tan x + \frac {b - \sqrt {b^2 - 4ac}} {2a}} {\tan x + \frac {b + \sqrt {b^2 - 4ac}} {2a}} \right| + C$$

where $b^2 - 4ac$ is positive which agrees with the answers I obtained with online calculators (wolframalpha and integral-calculator.)

In the case where $b^2 - 4ac = 0$, we have $c = \frac {b^2} {4a}$. Substituting c in (1), we have

$$\int \frac {du} {au^2 + bu + \frac {b^2} {4a}}$$

$$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \frac {b^2} {4a^2}}$$

$$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \left( \frac b {2a} \right)^2}$$

$$\frac 1 a \int \frac {du} {\left( u + \frac b {2a} \right)^2}$$

Let $v = u + \frac b {2a} \Rightarrow du = dv$

$$\frac 1 a \int \frac {dv} {v^2}$$

$$\frac {-1} {av} + D$$

$$\frac {-1} {a \left( u + \frac b {2a} \right)} + D$$

$$\frac {-2} {2au + b} + D$$

$$\frac {-2} {2a \tan x + b} + D$$

Is my solution for when $b^2 - 4ac = 0$ correct?

How do I go about evaluating the integral when $b^2 - 4ac < 0$? Looking at (1), the denominator can't be factored in this case. Trying Weiserstrass substitution didn't seem to work but maybe I didn't persist long enough.

Any feedback on my form or how I went about evaluating the integral would be appreciated.

 

[Charles Link]

When $b^2-4ac <0$, I believe you have the equivalent of $\int \frac{1}{x^2+1} \, dx=\tan^{-1}{x}+C$.

 

[Ray Vickson]

Homework Statement

If $a \neq 0$, evaluate the integral

$$\int \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$

(Hint: Make the substitution $u = \tan x$ and consider separately the cases where
$b^2 - 4ac$ is positive, zero, or negative.)

The Attempt at a Solution

How do I go about evaluating the integral when $b^2 - 4ac < 0$? Looking at (1), the denominator can't be factored in this case. Trying Weiserstrass substitution didn't seem to work but maybe I didn't persist long enough.

Any feedback on my form or how I went about evaluating the integral would be appreciated.

Let $k = b/(2a)$. If $(b^2-4ac)>0,$ let $A = \sqrt{b^2-4ac}/(2a)$, so (assuming $a>0$) your integral is
$$\int \frac{du}{(u+k)^2 - A^2}$$
with $A > 0.$
If $(b^2-4ac) < 0,$ let $B = \sqrt{4ac - b^2}/(2a),$ so (again, assuming $a>0$) your integral is
$$\int \frac{du}{(u+k)^2 + B^2}$$
with $B > 0.$

Both of these are standard, and can be found in integral tables, etc. Or, you can do them manually.