Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$

Question

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

Answer 1

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

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

Answer 2

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, $$2\cos{x}=5\mathrm{A}\cos{x}+3\mathrm{B}\cos{x}$$ $$-\sin{x}=3\mathrm{A}\sin{x}-5\mathrm{B}\sin{x}$$ $$ \mathrm{A}=\frac{7}{34} \; \text{and} \; \mathrm{B}=\frac{11}{34}$$ $$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\int \frac{7}{34}\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +\frac{11}{34} \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; \mathrm{d}x$$ $$\boxed{\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\frac{7x}{34}+\frac{11\ln{\big | 3\sin{x}+5\cos{x} \big |}}{34}+\mathrm{C}}$$

Answer 3

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.

Answer 4

hint

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

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

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