Mathematics

If the integral $$\int \dfrac {5 \tan x}{\tan x-2}dx=x+\alpha In|\sin x-2 \cos x|+k$$ , then $$a$$ is equal to


ANSWER

$$2$$


SOLUTION
$$I = \int\dfrac{5 \ tan x}{tan x -2}dx$$

$$I = \int\dfrac{5 \ \dfrac{sin \ x}{cos\ x}}{\dfrac{sin\ x }{cos\ x} -2}dx\Rightarrow \int\dfrac{5 sin x}{sinx-2cos x}dx$$

$$\begin{bmatrix}\because\displaystyle 5\  sin\ x\leftrightarrow sin\ x - 2\ cos x +2\ cos \ x+ 4\ sin\  x\end{bmatrix}$$


$$I = \int\begin{pmatrix}\dfrac{sin\ x - 2\ cos\  x}{sin\ x-2\ cos \ x}+\dfrac{4\ sin\ x+2\ cos\ x}{sin\ x-2\ cos\ x}\end{pmatrix}dx$$

$$I = \int\begin{pmatrix}\dfrac{sin\ x - 2\ cos\  x}{sin\ x-2\ cos \ x}\end{pmatrix}dx+\int\begin{pmatrix}\dfrac{4\ sin\ x+2\ cos\ x}{sin\ x-2\ cos\ x}\end{pmatrix}dx$$

$$I=I_1+I_2$$ ..............(i)

$$I_1 = \int dx \Rightarrow x+c_1$$

$$I_2=\int\dfrac{2\ (2\ sin \ x+cos\ x)}{sin\ x-2\ cos\  x}dx$$

put $$sin\ x-2\ cos\ x=t\rightarrow(2\ sin\  x+cos\  x)dx= dt$$

$$\Rightarrow \int\dfrac{2}{t}dt$$

$$\Rightarrow2\ ln\begin{vmatrix}t\end{vmatrix}+c_2$$
from (i) we get,

$$I=x+ 2\ ln\begin{vmatrix}sin\ x-2\ cos\ x\end{vmatrix}+ C$$........(ii)

$$I=x+ \alpha\ ln\begin{vmatrix}sin\ x-2\ cos\ x\end{vmatrix}+ C$$...........(iii)

from (ii) and (iii)

$$\alpha= 2$$

$$\therefore \text{correct option is D}$$
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