Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$

Question

Evaluate

$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$

My Try:

$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$

we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and

$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$

So

$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$

So

$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$

where

$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$

I got stuck up to integrate $J$.

Answer 1

What about the following? Note that the denominator can be written as $$3 \sin x + 4 \cos x = 5 \sin (x + \theta)$$ where $\theta$ is easy to determine. The integral becomes $$I= \frac{1}{5} \int \frac{\cos 2x \: dx}{\sin (x + \theta)}.$$

Now change the variable of integration from $x$ to $y = x + \theta$ so that you now have to evaluate $$I= \frac{1}{5} \int \frac{\cos (2y - 2 \theta) \: dy}{\sin y}$$ and use the expansion formula for $\cos (2 y - 2 \theta)$. Eventually you have to evaluate integrals of the form $ \int \frac{\cos 2 y}{\sin y} dy$ and $ \int \frac{\sin 2y}{\sin y} dy$ which are straight forward(?).

Answer 2

HINT: set $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ $$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ your Integrand is given by $$-1/2\,{\frac { \left( {t}^{2}+2\,t-1 \right) \left( {t}^{2}-2\,t-1 \right) }{ \left( 2\,t+1 \right) \left( t-2 \right) \left( {t}^{2}+ 1 \right) }} $$ with $t=\tan(x/2)$ and don't Forget $$dx=\frac{2dt}{1+t^2}$$

Answer 3

$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x} = \dfrac15\int \frac{\cos 2x \: dx}{\dfrac35 \sin x+\dfrac45 \cos x} = \dfrac15\int \frac{\cos 2x \: dx}{\cos(x - \alpha)}$$

Where $\alpha = \cos^{-1} \dfrac45$

$$\dfrac15\int \frac{\cos 2x \: dx}{\cos(x - \alpha)} = \dfrac15\int \frac{2\cos^2 x - 1 \: }{\cos(x - \alpha)}\ dx = \dfrac15\int \frac{2\cos^2 x \: }{\cos(x - \alpha)}\ dx - \dfrac15\int \sec(x - \alpha) dx$$

Now for the first integral, let $u = x - \alpha$

$$\dfrac25\int \frac{\cos^2 (u + \alpha) }{\cos(u)}\ du = \dfrac25\int \frac{(\cos u \cos \alpha - \sin u \sin \alpha)^2 }{\cos(u)}\ du \\= \dfrac25\int\frac{(\cos^2 u \cos^2 \alpha + \sin^2 u \sin^2 \alpha) + 2\sin \alpha \cos \alpha \sin u \cos u }{\cos(u)}\ du\\=\dfrac25\int{\cos u \cos^2 \alpha}\ du + \dfrac25\int \frac{\sin^2 u \sin^2 \alpha}{\cos u} du+ \dfrac25\int {2\sin \alpha \cos \alpha \sin u }{}\ du $$

The middle integral,

$$\dfrac25\int \frac{(1 -\cos^2 u) \sin^2 \alpha}{\cos u}\ du = \dfrac25\int{\sec u \sin^2 \alpha}\ du - \dfrac25\int \cos u \sin^2 \alpha \ du$$

So the answer is,

$$I = \dfrac25\int{\sec u \sin^2 \alpha}\ du - \dfrac25\int \cos u \sin^2 \alpha \ du + \dfrac25\int{\cos u \cos^2 \alpha}\ du + \dfrac25\int {2\sin \alpha \cos \alpha \sin u }{}\ du -\dfrac15\int \sec(u) du$$

$$I = \dfrac25\sin^2 \alpha\int{\sec u }\ du + \dfrac25\cos (2\alpha)\int{\cos u }\ du + \dfrac25\sin (2\alpha)\int { \sin u }{}\ du -\dfrac15\int \sec(u) du$$

Answer 4

Note that $$3\sin x+4\cos x=5\Big( \dfrac{3}{5}\sin x+\dfrac{4}{5}\cos x \Big)$$ and $$\cos2x=1-2\sin^2x$$ also $\arcsin(4/5)=\arccos(3/5)=0.92$

combining all this, your integral becomes $\frac{1}{5}\int \frac{1 -2\sin^2x\: dx}{\sin(x+0.92)}$.

From here, writing $x=(x+0.92)-0.92$ in numerator and basic integration formula for $\csc x$, $\cot x$ will work.

Answer 5

\begin{align} Hint:(3\cos x-4\sin x)^2&=25-(3\sin x+4\cos x)^2\\ \int\frac{(3\cos x-4\sin x)^2}{3\sin x+4\cos x}dx&=\int\frac{25-(3\sin x+4\cos x)^2}{3\sin x+4\cos x}dx\\ &=25\int\frac{dx}{3\sin x+4\cos x}-\int(3\sin x+4\cos x)dx\\ &=25\int\frac{dx}{3\sin x-4+4(1+\cos x)}+3\cos x-4\sin x\\ &=25\int\frac{\frac{1}{1+\cos x}dx}{3\frac{\sin x}{1+\cos x}-4\frac{1}{1+\cos x}+4}+3\cos x-4\sin x\\ &=25\int\frac{d(\frac{\sin x}{1+\cos x})}{3\frac{\sin x}{1+\cos x}-2((\frac{\sin x}{1+\cos x})^2+1)+4}+3\cos x-4\sin x\\ &=25\int\frac{dt}{2+3t-2t^2}+3\cos x-4\sin x\\ &=25(\frac{1}{5}\ln \left|\frac{2t+1}{2-t}\right |)+3\cos x-4\sin x\\ &=5\ln \left|\frac{2\sin x+1+\cos x}{2(1+\cos x)-\sin x}\right |+3\cos x-4\sin x+C\\ \end{align}