$\int \frac{5 \tan x}{(\tan x)-2} d x=\alpha x+\beta \log |\sin x-2 \cos x|$ $+\gamma$, then $\alpha-\beta=$
$\int \frac{5 \tan x}{(\tan x)-2} d x=\alpha x+\beta \log |\sin x-2 \cos x|$ $+\gamma$, then $\alpha-\beta=$
- –1
- 2
- 0
- 1
Solution
$\int \frac{5 \tan x}{\tan x-2} d x=\int \frac{5 \sin x}{\sin x-2 \cos x} d x$
Put, $5 \sin x=\alpha(\sin x-2 \cos x)+$
$
\begin{array}{r}
\beta \frac{d}{d x}(\sin x-2 \cos x) \\
5 \sin x=\alpha(\sin x-2 \cos x)+\beta(\cos x+2 \sin x)
\end{array}
$
On comparison of coefficients of $\sin x, \cos x$ we get
$\sin x$ coefficients $\rightarrow 5=\alpha+2 \beta$
$\cos x$ coefficients $\rightarrow 0=-2 \alpha+\beta$
On solving Eqs. (i) and (ii), weget
$
\begin{gathered}
\alpha=1 \text { and } \beta=2 \\
\therefore \quad 5 \sin x=1(\sin x-2 \cos x)+2(\cos x+2 \sin x)
\end{gathered}
$
Now, substitute $5 \sin x$ value in the given question.
$
\begin{aligned}
& \int \frac{5 \sin x}{(\sin x-2 \cos x)} d x \\
& \quad=\int \frac{1(\sin x-2 \cos x)+2(\cos x+2 \sin x)}{(\sin x-2 \cos x)} d x \\
& \quad=\int \frac{1(\sin x-2 \cos x)}{\sin x-2 \cos x} d x+2 \int \frac{\cos x+2 \sin x}{(\sin x-2 \cos x)} d x \\
& \quad=\int 1 \cdot d x+2 \log |\sin x-2 \cos x|+c \\
& \quad\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right] \\
& =x+2 \log |\sin x-2 \cos x|+c
\end{aligned}
$
On comparison with given RHS, we get
$
\alpha=1, \beta=2 \Rightarrow \alpha-\beta=-1
$
Hence, option (1) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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