\(\int\cos^3x\sin^2x\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle \frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C\)
The power of \(\cos x\) is odd, so one \(\cos x\) is kept for \(dt\) and the rest is written in \(\sin x\):
\( \cos^3x\sin^2x = \left(1-\sin^2x\right)\sin^2x\cos x \)
With the substitution \(t = \sin x\), \(dt = \cos x\,dx\):
\( \int\left(1-t^2\right)t^2\,dt = \int\left(t^2-t^4\right)dt \)
\( \displaystyle = \frac{t^3}{3}-\frac{t^5}{5}+C \)
\( \displaystyle = \frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C \)
Why the other options are incorrect
Option (2) \(\frac{\sin^3x}{3}+\frac{\sin^5x}{5}+C\) — Incorrect. It uses \(\cos^2x = 1+\sin^2x\); the identity gives \(1-\sin^2x\).
Option (3) \(\frac{\sin^5x}{5}-\frac{\sin^3x}{3}+C\) — Incorrect. It is the negative of the answer; its derivative is \(-\cos^3x\sin^2x\).
Option (4) \(\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C\) — Incorrect. This is the integral of \(\sin^3x\cos^2x\), with the two powers the other way round.