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$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to
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Verified Answer
The correct answer is:
$x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c$
Let $I=\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$
We can write
$\sin x+8 \cos x=A(4 \sin x+6 \cos x)$
$+B \frac{d}{d x}(4 \sin x+6 \cos x)$
$\sin x+8 \cos x$
$=A(4 \sin x+6 \cos x)+B(4 \cos x-6 \sin x)$
On equating the coefficient of $\sin x$ and $\cos x$, we get
$\begin{array}{ll}1=4 A-6 B, & 8=6 A+4 B \\ \Rightarrow & A=1, B=\frac{1}{2}\end{array}$
$\begin{aligned} \therefore I & =\int \frac{(4 \sin x+6 \cos x)+\frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x \\ & =\int\left(1+\frac{1}{2} \cdot \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x}\right) d x \\ & =x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c\end{aligned}$
We can write
$\sin x+8 \cos x=A(4 \sin x+6 \cos x)$
$+B \frac{d}{d x}(4 \sin x+6 \cos x)$
$\sin x+8 \cos x$
$=A(4 \sin x+6 \cos x)+B(4 \cos x-6 \sin x)$
On equating the coefficient of $\sin x$ and $\cos x$, we get
$\begin{array}{ll}1=4 A-6 B, & 8=6 A+4 B \\ \Rightarrow & A=1, B=\frac{1}{2}\end{array}$
$\begin{aligned} \therefore I & =\int \frac{(4 \sin x+6 \cos x)+\frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x \\ & =\int\left(1+\frac{1}{2} \cdot \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x}\right) d x \\ & =x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c\end{aligned}$
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