Question

Evaluate :
$\int \frac{2\cos x+3\sin x}{4\cos x+5\sin x}dx.$

Answer 1

$\int \frac{2\cos x+3\sin x}{4\cos x+5\sin x}dx$

$=\int \frac{4\cos x+5\sin x-2(\cos x+\sin x)}{4\cos x+5\sin x}dx$

$=\int \frac{4\cos x+5\sin x}{4\cos x+5\sin x}dx-\int \frac{2\cos x+2\sin x}{4\cos x+5\sin x}dx$

$=\int dx+\frac{1}{2}\int \frac{-4\cos x-4\sin x+5\cos x-5\cos x}{4\cos x+5\sin x}$

$=\int dx+\frac{1}{2}\int \frac{5\cos x-4\sin x}{4\cos x+5\sin x}dx-\frac{9}{2}\int \frac{\sin x}{4\cos x+5\sin x}dx.....\left(1\right)$

Now,

$\int \frac{5\cos x-4\sin x}{4\cos x+5\sin x}dx$

Let,

$4\cos x+5\sin x=t$

$(5\cos x-4\sin x)dx=dt$

Therefore, the integral become

$\int \frac{dt}{t}=\log t+C_1$

Substituting the value of $t$ in above equation, we get

$\int \frac{5\cos x-4\sin x}{4\cos x+5\sin x}dx=\log (4\cos x+5\sin x)+C_1.....\left(2\right)$

Now,

$\int \frac{\sin x}{4\cos x+5\sin x}dx$

$=\frac{1}{\sqrt{41}}\int \frac{\sin x}{\frac{4}{\sqrt{41}}\cos x+\frac{5}{\sqrt{41}}\sin x}dx$

$=\frac{1}{\sqrt{41}}\int \frac{\sin x}{\sin \alpha \cos x+\cos \alpha \sin x}dx[\text{Here}\sin \alpha =\frac{4}{\sqrt{41}}\text{and}\cos \alpha =\frac{5}{\sqrt{41}}]$

$=\frac{1}{\sqrt{41}}\int \frac{\sin xdx}{\sin (x+\alpha )}$

Let $x+\alpha =t\Rightarrow x=t-\alpha \Rightarrow dx=dt$

Therefore, the integral become

$\int \frac{\sin (t-\alpha )}{\sin t}dt$

$=\int \frac{\sin t\cos \alpha -\cos t\sin \alpha }{\sin t}dt$

$=\int \frac{\sin t\cos \alpha }{\sin t}dt-\int \frac{\cos t\sin \alpha }{\sin t}dt$

$=\int \cos \alpha dt-\int \cot t\sin \alpha dt$

$=t\cos \alpha -\sin \alpha \log \left(\sin t\right)+C_2$

substituting the valye of $t$ in above equation we get

$\int \frac{\sin x}{4\cos x+5\sin x}dx$ $=(x+\alpha )\cos \alpha -\sin \alpha \ln \left(\sin (x+\alpha )\right)+C_2=(x+{\tan }^{-1}\left(\frac{4}{5}\right))\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}\ln \left(\sin (x+{\tan }^{-1}\frac{4}{5})\right)+C_2.....\left(3\right)$

From equation $\left(1\right),\left(2\right)\&\left(3\right)$, we have

$\int \frac{2\cos x+3\sin x}{4\cos x+5\sin x}dx$

$=x+\frac{1}{2}\log (4\cos x+5\sin x)-\frac{9}{2}[(x+{\tan }^{-1}\left(\frac{4}{5}\right))\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}\ln \left(\sin (x+{\tan }^{-1}\frac{4}{5})\right)]+C$