Question:
\(\displaystyle \int \frac{3\cos x-2\sin x}{2\cos x+3\sin x}\,dx=\ \ ?\)
\(\displaystyle \int \frac{3\cos x-2\sin x}{2\cos x+3\sin x}\,dx=\ \ ?\)
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Spot "derivative over itself" $\Rightarrow$ immediate $\ln|\,\cdot\,|$.
Updated On: Oct 17, 2025
- \(2\cos x+3\sin x+k\)
- \(\log|2\cos x+3\sin x|+k\)
- \(\tan^{-1}\!\big(3\sin\tfrac{x}{2}\big)+k\)
- \(2\tan\tfrac{x}{2}+k\)
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The Correct Option is B
Solution and Explanation
Let \(f(x)=2\cos x+3\sin x\). Then
\[
f'(x)=-2\sin x+3\cos x=3\cos x-2\sin x,
\]
which is exactly the numerator.
Hence the integrand is \(\dfrac{f'(x)}{f(x)}\), so
\(\int \dfrac{f'}{f}\,dx=\ln|f|+k=\ln|2\cos x+3\sin x|+k\).
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