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If $\int \frac{5 \tan (x)}{\tan (x)-2} d x=x+a \log$
$|\sin (x)-2 \cos (x)|+k$, then $a$ is equal to
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$|\sin (x)-2 \cos (x)|+k$, then $a$ is equal to
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1812 Upvotes
Verified Answer
The correct answer is:
2
We have,
$$
\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+K
$$
On differentiating, we get
$$
\begin{aligned}
& \frac{5 \tan x}{\tan x-2}=1+\frac{a(\cos x+2 \sin x)}{\sin x-2 \cos x} \\
\Rightarrow \quad \frac{5 \tan x}{\tan x-2} & =1+\frac{a(1+2 \tan x)}{\tan x-2} \\
\Rightarrow \quad & \frac{5 \tan x}{\tan x-2}=\frac{\tan x-2+a+2 a \tan x}{\tan x-2} \\
\Rightarrow \quad 5 \tan x & =(2 a+1) \tan x+a-2
\end{aligned}
$$
Equating the coefficient of $\tan x$ and constant terms, we get
$$
\begin{aligned}
2 a+1 & =5 \\
a-2 & =0 \\
\therefore \quad a & =2
\end{aligned}
$$
$$
\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+K
$$
On differentiating, we get
$$
\begin{aligned}
& \frac{5 \tan x}{\tan x-2}=1+\frac{a(\cos x+2 \sin x)}{\sin x-2 \cos x} \\
\Rightarrow \quad \frac{5 \tan x}{\tan x-2} & =1+\frac{a(1+2 \tan x)}{\tan x-2} \\
\Rightarrow \quad & \frac{5 \tan x}{\tan x-2}=\frac{\tan x-2+a+2 a \tan x}{\tan x-2} \\
\Rightarrow \quad 5 \tan x & =(2 a+1) \tan x+a-2
\end{aligned}
$$
Equating the coefficient of $\tan x$ and constant terms, we get
$$
\begin{aligned}
2 a+1 & =5 \\
a-2 & =0 \\
\therefore \quad a & =2
\end{aligned}
$$
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