\(\int\cos 6x\sqrt{1+\sin 6x}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{9}(1+\sin 6x)^{3/2}+C\)
The factor \(\cos 6x\) is, up to a constant, the derivative of \(1+\sin 6x\), so substitute
\(t = 1+\sin 6x,\quad dt = 6\cos 6x\,dx\)
Then
\( \displaystyle \int\cos 6x\sqrt{1+\sin 6x}\,dx = \frac{1}{6}\int t^{1/2}\,dt \)
\( \displaystyle = \frac{1}{6}\times\frac{2}{3}t^{3/2}+C \)
\( \displaystyle = \frac{1}{9}(1+\sin 6x)^{3/2}+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{6}(1+\sin 6x)^{3/2}+C\) — Incorrect. It forgets the factor \(\frac{2}{3}\) from \(\int t^{1/2}\,dt = \frac{2}{3}t^{3/2}\).
Option (2) \(\displaystyle\frac{2}{9}(1+\sin 6x)^{3/2}+C\) — Incorrect. Its derivative is \(2\cos 6x\sqrt{1+\sin 6x}\), twice the integrand.
Option (3) \(\displaystyle\frac{1}{9}(1+\cos 6x)^{3/2}+C\) — Incorrect. The expression under the root is \(1+\sin 6x\), not \(1+\cos 6x\).