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

In summary: You will need to use partial fractions in both cases.In summary, to evaluate the integral $$\int \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x},$$ you can make the substitution ##u = \tan x## and consider three cases: when ##b^2 - 4ac > 0, b^2 - 4ac = 0,## and ##b^2 - 4ac < 0.## In the first two cases, you will need to use partial fractions, and in the third case, you can use the standard integral $$\int \frac {dx} {x
  • #1
Entertainment Unit
16
1

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.
 
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  • #2
When ## b^2-4ac <0 ##, I believe you have the equivalent of ## \int \frac{1}{x^2+1} \, dx=\tan^{-1}{x}+C ##.
:welcome:
 
  • #3
Entertainment Unit said:

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.
 
Last edited:

What is the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x)?

The integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) cannot be expressed in terms of elementary functions. It is a special type of integral known as an elliptic integral, which can only be solved using advanced mathematical techniques.

Why is the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) important?

The integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) has applications in various fields of science and engineering, particularly in physics and mathematics. It is used to solve problems involving oscillatory systems, such as in optics, mechanics, and electromagnetic theory.

Can the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) be approximated?

Yes, the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) can be approximated numerically using various techniques, such as the trapezoidal rule, Simpson's rule, or the Monte Carlo method. However, the accuracy of the approximation depends on the values of a, b, and c.

How is the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) related to other mathematical concepts?

The integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) is closely related to other mathematical concepts, such as trigonometric functions, complex numbers, and differential equations. It also has connections to elliptic curves and modular forms, which are important in modern number theory and cryptography.

Are there any special cases where the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) can be solved analytically?

Yes, there are special cases where the integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x) can be solved analytically. For example, when a = b = c = 1, the integral simplifies to 1/x, which can be easily integrated. Additionally, when a = 0 or c = 0, the integral can be reduced to a combination of elementary functions.

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