4
$\begingroup$

How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$

I tried using the half angle formula $$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$

substituted and simplified I got

$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$ substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$ $$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$ I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance

| cite | improve this question | |
$\endgroup$
8
$\begingroup$

The trick that I use for solving integrals with $\sin{x}$ and $\cos{x}$ with varying coefficients in the numerator and denominator as follows.

For your integral, consider the two easy integrals: \begin{align*} \int \frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x&=x+\mathrm{C}\\ \int \frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x&=\ln{\big | 3\sin{x}+5\cos{x} \big |}+\mathrm{C}\\ \end{align*} Now, set up a system of equations as the following: $$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x=\int \mathrm{A}\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +\mathrm{B} \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; \mathrm{d}x$$ Such that, $$2\cos{x}=5\mathrm{A}\cos{x}+3\mathrm{B}\cos{x}$$ $$-\sin{x}=3\mathrm{A}\sin{x}-5\mathrm{B}\sin{x}$$ $$ \mathrm{A}=\frac{7}{34} \; \text{and} \; \mathrm{B}=\frac{11}{34}$$ $$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\int \frac{7}{34}\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +\frac{11}{34} \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; \mathrm{d}x$$ $$\boxed{\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\frac{7x}{34}+\frac{11\ln{\big | 3\sin{x}+5\cos{x} \big |}}{34}+\mathrm{C}}$$

| cite | improve this answer | |
$\endgroup$
  • 5
    $\begingroup$ Very beautiful...in the $\ln$ missing the absolute value. $\endgroup$ – Sebastiano Jun 24 at 22:41
  • 2
    $\begingroup$ @Sebastiano Thank you. I fixed it. $\endgroup$ – Ty. Jun 24 at 22:42
  • 3
    $\begingroup$ a coefficient $\frac{11}{34}$ is also missing $\endgroup$ – user766881 Jun 24 at 22:45
  • $\begingroup$ Hi, I have added also a question for your answer: math.stackexchange.com/questions/3734680/… $\endgroup$ – Sebastiano Jun 25 at 22:44
2
$\begingroup$

Bioche's rules say you should use the substitution $$t=\tan x,\qquad\mathrm dx=\frac{\mathrm dt}{1+t^2},$$ to obtain the integral of a rational function in $t$, which you can compute via partial fractions decomposition.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Never I have heared the Bioche's rules :-) What is? :-) $\endgroup$ – Sebastiano Jun 24 at 22:40
  • 1
    $\begingroup$ It is for the integral of arational function of trigonometric functions. It says the substitution is determined as follows: consider the differential form $\omega(x)=f(t)\,\mathrm dx$. If $\omega(-x)=f(-x)\,\mathrm d(-x) = \omega (x)$, then set $t=\cos x$. If $\omega(\pi-x)=\omega(x)$, set $t=\sin x$. If $\omega(t+\pi)=\omega(t)$, set $t=\tan x$. If, unfortunately, none works, you always the duplication formulæ, and can set $t=\tan\frac x2$. If two work, you can set $t=$ some trigonometric function of 2x$. You can see Bioche's rules on Wikipedia, or if you read French, Règles de Bioche . $\endgroup$ – Bernard Jun 24 at 22:55
  • $\begingroup$ Thank youuuuu.. I thinked to the briosches :-)))) $\endgroup$ – Sebastiano Jun 24 at 22:56
  • $\begingroup$ A small trigonometric integral for Sunday breakfast? :-) $\endgroup$ – Bernard Jun 24 at 23:00
  • $\begingroup$ No no...please..I must detox from mathematics and physics :-)))))) $\endgroup$ – Sebastiano Jun 24 at 23:02
2
$\begingroup$

Rewrite numerator: $2\cos x-\sin x=\frac{7}{34}(3\sin x+5\cos x)+\frac{11}{34}\frac{d}{dx}(3\sin x+5\cos x)$

$$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x }$$ $$=\int\dfrac{\frac{7}{34}\left(3\sin x+5\cos x\right)+\frac{11}{34}\left(3\cos x-5\sin x\right)}{3\sin x+5\cos x }dx$$ $$=\frac{7}{34}\int\ dx+\frac{11}{34}\int \frac{3\cos x-5\sin x}{3\sin x+5\cos x }dx$$ $$=\frac{7}{34}\int\ dx+\frac{11}{34}\int \frac{d(3\sin x+5\cos x)}{3\sin x+5\cos x }$$ $$=\frac{7x}{34}+\frac{11}{34}\ln|3\sin x+5\cos x|+C$$

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ I don't know why so many downvotes. I don't know what is wrong with this answer! $\endgroup$ – Harish Chandra Rajpoot Jun 24 at 22:41
  • 1
    $\begingroup$ For my humble opinion: nothing. For me it is not very clear, for example, $2\cos x-\sin x=A(3\sin x+5\cos x)+B\frac{d}{dx}(3\sin x+5\cos x)$. However when I have seen the -2 I have adjusted with +1 by me. $\endgroup$ – Sebastiano Jun 24 at 22:46
  • 2
    $\begingroup$ @Sebastiano Thanks. I adjusted the numerator as the sum of its denominator and derivative of denominator. May be it created confusion. Thanks for your suggestion. $\endgroup$ – Harish Chandra Rajpoot Jun 24 at 23:01
  • 1
    $\begingroup$ It is a pleasure for me to give suggestions and positive votes. I still don't understand why you got negative votes. I always wish you all the best. $\endgroup$ – Sebastiano Jun 24 at 23:03
2
$\begingroup$

hint

Dividing by $\cos(x)$ and putting $\tan(x)=t$, it becomes $$\int \frac{(2-t)dt}{(5+3t)(1+t^2)}$$

$$=\frac{1}{34}\int \Bigl(\frac{33}{5+3t}+\frac{-11t+7}{t^2+1}\Bigr)dt$$

$$=\frac{11}{34}\int \frac{3dt}{5+3t}-\frac{11}{68}\int \frac{2tdt}{t^2+1}+$$ $$\frac{7}{34}\int \frac{dt}{t^2+1}$$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ i think it further needs partial fractions? $\endgroup$ – user766881 Jun 24 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy