Question

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

Answer 1

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

$=\int \frac{dx}{5(\frac{4}{5}\cos x+\frac{3}{5}\sin x)}$

$=\frac{1}{5}\int \frac{dx}{\sin \alpha \cos x+\cos \alpha \sin x}$

Here, $\sin \alpha =\frac{4}{5};\cos \alpha =\frac{3}{5}$

$=\frac{1}{5}\int \frac{dx}{\sin (\alpha +x)}$

$=\frac{1}{5}\ln \left|\tan \left(\frac{\alpha +x}{2}\right)\right|+C.....\left(1\right)$

Now, solving for $\tan \left(\frac{\alpha +x}{2}\right)$

$\tan \left(\frac{\alpha +x}{2}\right)$

$=\frac{\tan \frac{\alpha }{2}+\tan \frac{x}{2}}{1-\tan \frac{\alpha }{2}\tan \frac{x}{2}}$

$=\frac{\frac{\sin \alpha }{1+\cos \alpha }+\left(\frac{\sin x}{1+\cos x}\right)}{1-\left(\frac{\sin \alpha }{1+\cos \alpha }\right)\left(\frac{\sin x}{1+\cos x}\right)}$

$=\frac{1+2\tan \frac{x}{2}}{2-\tan \frac{x}{2}}$

$=\frac{1+\cos x+2\sin x}{2+2\cos x-\sin x}$

Substituting the above value in equation $\left(1\right)$, we get

$\int \frac{dx}{4\cos x+3\sin x}=\frac{1}{5}\ln \left|\frac{1+\cos x+2\sin x}{2+2\cos x-\sin x}\right|+C$