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AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Section: Mathematics

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Question : 40 of 160

Marks: +1, -0

\int \frac{5\tan x}{(\tan x)-2}

dx

\alpha x+\beta\log|\sin x-2\cos x|+\gamma

\alpha-\beta=

$-1$
$2$
$0$
$1$

Solution:

\int\;\frac{5\tan x}{\tan x-2}dx = \int\;\frac{5\sin x}{\sin x-2\cos x}dx

Put,

5\sin x=\alpha(\sin x-2\cos x)+

\beta\;\frac{d}{dx}(\sin x-2\cos x)

5\sin x=\alpha(\sin x-2\cos x)+\beta(\cos x+2\sin x)

On comparison of coefficients of

\sin x, \cos x

we get

\sin x

coefficients

\rightarrow 5=\alpha+2\beta

...(i)

\cos x

coefficients

\rightarrow 0=-2\alpha+\beta

On solving Eqs. (i) and (ii), weget

\alpha=1

and

\beta=2

\therefore \;\; 5\sin x=1(\sin x-2\cos x)+2(\cos x+2\sin x)

Now, substitute

5\sin x

value in the given question.

\int\;\frac{5\sin x}{(\sin x-2\cos x)}dx

=\int\;\frac{1(\sin x-2\cos x)+2(\cos x+2\sin x)}{(\sin x-2\cos x)}dx

=\int\;\frac{1(\sin x-2\cos x)}{\sin x-2\cos x}dx + 2\int\;\frac{\cos x+2\sin x}{(\sin x-2\cos x)}dx

=\int 1 \cdot dx + 2\log|\sin x-2\cos x|+c

\left| \because \int\;\frac{f'(x)}{f(x)}dx = \log|f(x)| \right|

=x+2\log|\sin x-2\cos x|+c

On comparison with given RHS, we get

\alpha=1, \beta=2 \Rightarrow \alpha-\beta=-1

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