Question

integration of
sin3x cos^2x​ dx

Answer 1

Answer:

math]\begin{align}I&=\int\sin\ 3x \cdot \cos\ 2x\ dx\\&=\frac{1}{2}\int(\sin\ \underbrace{5x}_{\alpha}+\sin\ \underbrace{x}_{\beta})\ dx \qquad \left[\sin\alpha+\sin\beta=2\sin\ \frac{\alpha+\beta}{2}.\cos\ \frac{\alpha-\beta}{2}\right]\\&=\frac{1}{2}\left\{\underbrace{\int(\sin\ 5x\ dx}_{-\frac{1}{5}\cos\ 5x}+\underbrace{\int(\sin\ x\ dx}_{-\cos\ x}\right\}\\&=\boxed{-\frac{1}{10}\cos\ 5x--\frac{1}{2}\cos\ x+C}\end{align}\tag*{}[/math]

Step-by-step explanation:

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