Question

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

• A. $\frac{10}{29}x-\frac{16}{29}\log \left(2\cos x+5\sin x\right)+c$
• B. $\frac{9}{29}x-\frac{25}{29}\log \left(2\cos x+5\sin x\right)+c$
• C. $\frac{7}{29}x-\frac{26}{29}\log \left(2\cos x+5\sin x\right)+c$
• D. none of these

Answer 1

The correct option is C $\frac{7}{29}x-\frac{26}{29}\log \left(2\cos x+5\sin x\right)+c$

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

$3\sin x-4\cos x=a(2\cos x+5\sin x)+b\frac{d}{dx}(2\cos x+5\sin x)$

$\Rightarrow 3\sin x-4\cos x=a(2\cos x+5\sin x)+b(-2\sin x+5\cos x)$

$\Rightarrow 3\sin x-4\cos x=(5a-2b)\sin x+(2a+5b)\cos x$

$\Rightarrow (5a-2b-3)\sin x+(2a+5b+4)\cos x=0$

$\Rightarrow 5a-2b-3=0,2a+5b+4=0$

$\Rightarrow 5a-2b=3$ ..........$\left(1\right)$

$\Rightarrow 2a+5b=-4$ ..........$\left(2\right)$

$2\times equation\left(1\right)-5\times equation\left(2\right)$

$\Rightarrow 10a-4b-10a-25b=6+20$

$\Rightarrow -29b=26$

$\therefore b=-\frac{26}{29}$

Substitute $b=-\frac{26}{29}$ in $\left(1\right)$ we get

$5a-2b=3$

$\Rightarrow 5a=3+2b=3+2\times -\frac{26}{29}=\frac{87-52}{29}=\frac{35}{29}$

$\Rightarrow a=\frac{7}{29}$

$\therefore a=\frac{7}{29}$ and $b=-\frac{26}{29}$

$\therefore I=\int \frac{3\sin x-4\cos x}{2\cos x+5\sin x}dx$

$=\frac{7}{29}\int \frac{2\cos x+5\cos x}{2\cos x+5\sin x}dx-\frac{26}{29}\int \frac{-2\sin x+5\cos x}{2\cos x+5\sin x}dx$

$=\frac{7}{29}\int dx-\frac{26}{29}\int \frac{-2\sin x+5\cos x}{2\cos x+5\sin x}dx$

$=\frac{7}{29}x-\frac{26}{29}\ln |2\cos x+5\sin x|+c$