$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to
$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to
- $x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c$
- $2 x+\log (2 \sin x+3 \cos x)+c$
- $x+2 \log (2 \sin x+3 \cos x)+c$
- $\frac{1}{2} \log (4 \sin x+6 \cos x)+c$
Solution
\(\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x\)
Write numerator as:
\(\sin x+8 \cos x=\frac{1}{2}(4 \cos x-6 \sin x)+(4 \sin x+6 \cos x)\)
So,
\(\begin{gathered}
\int \frac{4 \sin x+6 \cos x}{4 \sin x+6 \cos x} d x+\frac{1}{2} \int \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x} d x \\
=\int d x+\frac{1}{2} \int \frac{d(4 \sin x+6 \cos x)}{4 \sin x+6 \cos x} \\
=x+\frac{1}{2} \log (4 \sin x+6 \cos x)+C
\end{gathered}\)
Asked in: MHT CET Full Test 13
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