Question

\(\displaystyle \int \frac{3\cos x-2\sin x}{2\cos x+3\sin x}\,dx=\ \ ?\)

• A. \(2\cos x+3\sin x+k\)
• B. \(\log|2\cos x+3\sin x|+k\)
• C. \(\tan^{-1}\!\big(3\sin\tfrac{x}{2}\big)+k\)
• D. \(2\tan\tfrac{x}{2}+k\)

Hint:

Spot "derivative over itself" $\Rightarrow$ immediate $\ln|\,\cdot\,|$.

Answer 1

The Correct Option is B

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 \(\int \dfrac{f'}{f}\,dx=\ln|f|+k=\ln|2\cos x+3\sin x|+k\).