Question

What is the value of \[\int\sin^3x\cos^2x\,dx\]?

Options

• \[\frac{1}{3}\cos^3x-\frac{1}{5}\cos^5x+C\]

• \[-\frac{1}{3}\cos^3x+\frac{1}{5}\cos^5x+C\]

• \[-\frac{1}{3}\sin^3x+\frac{1}{5}\sin^5x+C\]

• \[\frac{1}{3}\cos^3x+\frac{1}{5}\cos^5x+C\]

Answer 1

The transformed integral is \[-\int(t^2-t^4)\,dt\]. Integrating and substituting \[t=\cos x\] gives the stated result.