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How can I compute this integral?

$\int_{0}^{1} \sin^{2}(\pi x)\sin(m\pi x)\mathrm dx$

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2 Answers 2

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hint : first use $\sin^2 a=\frac{1}{2}(1-\cos2a)$ $$\begin{align}\int_{0}^{1} \sin^{2}(\pi x)\sin(m\pi x)\ dx&= \frac{1}{2}\int_{0}^{1} (1-\cos(\pi x))\sin(m\pi x)\ dx\\&= \frac{1}{2}\int_{0}^{1} (\sin(m\pi x)-\cos(\pi x)\sin(m\pi x))\ dx \end{align}$$

then use product to sum formula

$$ \begin{align} &=\frac{1}{2}\int_{0}^{1} \sin(m\pi x)\ dx-\frac{1}{2}\int_{0}^{1}\cos(\pi x)\sin(m\pi x)\ dx\\ &=\frac{1}{2}\int_{0}^{1} \sin(m\pi x)\ dx-\frac{1}{2}\times\frac{1}{2}\int_{0}^{1}(\sin(m\pi x+\pi x)+\sin(m\pi x-\pi x))\ dx\end{align}$$

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Hint

$$\sin^2(X)=\frac{1-\cos(2X)}{2}$$

$2\cos(a)\sin(b)=$

$\sin(b+a)+\sin(b-a)$.