If $\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x=\left(\frac{23}{41}\right) x+K \log$ $|4 \cos x+5…
If $\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x=\left(\frac{23}{41}\right) x+K \log$
$|4 \cos x+5 \sin x|+c$, then $K$ is equal to
- $\frac{2}{41}$
- $\frac{-2}{41}$
- $\frac{3}{41}$
- $\frac{-3}{41}$
Solution
We have,
$
\begin{aligned}
\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} & =\frac{23}{41} x \\
& +K \log |4 \cos x+5 \sin x|+C
\end{aligned}
$
On differentiating, we get
$
\begin{aligned}
& \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x}=\frac{23}{41}+\frac{K(-4 \sin x+5 \cos x)}{4 \cos x+5 \sin x} \\
& 2 \cos x+3 \sin x \\
&= \frac{92 \cos x+115 \sin x-164 K \sin x+205 K \cos x}{41} \\
& \Rightarrow 2 \cos x+3 \sin x=\left(\frac{92+205 K}{41}\right) \cos x \\
&+\left(\frac{115-164 K}{41}\right) \sin x
\end{aligned}
$
Equating the coefficient of $\cos x$ and $\sin x$, we get
$
\begin{aligned}
\frac{92+205 K}{41} & =2 \text { and } \frac{115-164 K}{41}=3 \\
\Rightarrow \quad 92+205 K & =82 \\
K & =\frac{-10}{205}=\frac{-2}{41}
\end{aligned}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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