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

## 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}##

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

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

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

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

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

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

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

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

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

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.

Homework Helper
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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
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## 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.

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