\(\int\displaystyle\frac{\left(2\sin x+1\right)\cos x}{10-\cos^2x-6\sin x}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle 2\log\left(3-\sin x\right)+\frac{7}{3-\sin x}+C\)
Writing \(\cos^2x = 1-\sin^2x\), the denominator is a perfect square in \(\sin x\):
\( 10-\cos^2x-6\sin x = \sin^2x-6\sin x+9 \)
\( = \left(3-\sin x\right)^2 \)
Let \(t = \sin x\), so \(dt = \cos x\,dx\):
\( \displaystyle I = \int\frac{2t+1}{(t-3)^2}\,dt \)
Since \(2t+1 = 2(t-3)+7\), the partial fractions are
\( \displaystyle \frac{2t+1}{(t-3)^2} = \frac{2}{t-3}+\frac{7}{(t-3)^2} \)
\( \displaystyle I = 2\log\left|t-3\right|-\frac{7}{t-3}+C \)
As \(t-3 = \sin x-3 = -\left(3-\sin x\right)\) and \(3-\sin x \gt 0\):
\( \displaystyle I = 2\log\left(3-\sin x\right)+\frac{7}{3-\sin x}+C \)
Why the other options are incorrect
Option (1) \(2\log\left(3-\sin x\right)-\frac{7}{3-\sin x}+C\) — Incorrect. It writes \(3-\sin x\) for \(t-3\) without changing the sign; \(-\frac{7}{t-3} = +\frac{7}{3-\sin x}\).
Option (2) \(2\log\left(3-\sin x\right)+\frac{5}{3-\sin x}+C\) — Incorrect. The constant left over is \(2t+1-2(t-3) = 7\), not 5.
Option (3) \(7\log\left(3-\sin x\right)+\frac{2}{3-\sin x}+C\) — Incorrect. The two coefficients are swapped; the log term carries 2.