Question

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

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$\frac{5tanx}{tanx-2}=\frac{5sinx}{sinx-2cosx}$

$\frac{d(sinx-2cosx)}{dx}=cosx+2sinx$

$5sinx=(sinx-2cosx)+2(cosx+2sinx)$

$\int \frac{5sinx}{sinx-2cosx}dx=\int \frac{(sinx-2cosx)+2(cosx+2sinx)}{sinx-2cosx}dx$

$=x+\int \frac{\left(2\right(cosx+2sinx)}{sinx-2cosx}dx$

$takingsinx-2cosx=t$

$=x+\int 2\left(1/t\right)dt=x+2\ln t+k=x+2\ln \left|sinx-2cosx\right|+k$

$\therefore a=2$