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Calculus and Beyond Homework Help
Integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x)
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[QUOTE="Entertainment Unit, post: 6043172, member: 649715"] [h2]Homework Statement [/h2] 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.) [h2]The Attempt at a Solution[/h2] $$\int \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$[B] [/B] $$\int \frac {\sec^2~x} {\sec^2~x} \frac {dx} {a~\sin^2~x + b~\sin~x~\cos~x + c~\cos^2~x}$$ [B][/B] $$\int \frac {\sec^2~x~dx} {a~\tan^2~x + b~\tan~x + c}$$ Let ##u = \tan~x \Rightarrow dx = \frac {du} {\sec^2~x}## [B][/B] $$\int \frac {sec^2~x} {au^2 + bu + c} \frac {du} {\sec^2~x}$$ $$\begin{equation} \tag{1} \int \frac {du} {au^2 + bu + c} \end{equation}$$ [B][/B] $$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \frac c a}$$ $$\frac 1 a \int \frac {du} {u^2 + \frac b a u + \left( \frac b {2a} \right)^2 - \left(\frac b {2a} \right)^2 + \frac c a}$$ $$\frac 1 a \int \frac {du} { \left (u + \frac b {2a} \right)^2 - \frac {b^2 - 4ac} {(2a)^2}}$$ $$\frac 1 a \int \frac {du} { \left(u + \frac b {2a} \right)^2 - \left (\frac {\sqrt {b^2 - 4ac}} {2a} \right)^2}$$ $$\frac 1 a \frac 1 {\frac {2 \sqrt {b^2 - 4ac} } {2a}} \ln \left| \frac {u + \frac b {2a} - \frac {\sqrt {b^2 - 4ac}} {2a}} {u + \frac b {2a} + \frac {\sqrt {b^2 - 4ac}} {2a}} \right| + C$$ $$\frac 1 {\sqrt {b^2 - 4ac}} \ln \left| \frac {u + \frac {b - \sqrt {b^2 - 4ac}} {2a}} {u + \frac {b + \sqrt {b^2 - 4ac}} {2a}} \right| + C$$ $$\frac 1 {\sqrt {b^2 - 4ac}} \ln \left| \frac {\tan x + \frac {b - \sqrt {b^2 - 4ac}} {2a}} {\tan x + \frac {b + \sqrt {b^2 - 4ac}} {2a}} \right| + C$$ 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. [/QUOTE]
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Integral of 1/(a sin^2 x + b sin x cos x + c cos^2 x)
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