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I'm trying to compute the following Integral $\int_{0}^{\pi} {\cos(x)\sin(2x)}dx$

This is what i've got so far:

$\int_{0}^{\pi} {\cos(x)\sin(2x)}dx =\int_{0}^{\pi} {\cos(x)2\sin(x)\cos(x)}dx = 2\int_{0}^{\pi} {\cos^2(x)\sin(x)}dx =$ $ 2 \int_{0}^{\pi} {\sin(x)(1-\sin^2(x))}dx = 2 \int_{0}^{\pi} {\sin(x)-\sin^3(x)}dx = 2 \int_{0}^{\pi} {\sin(x)} - 2 \int_{0}^{\pi}{\sin(x)^3}dx$

$ 2 \int_{0}^{\pi} {\sin(x)} = 2 [-\cos(x)]_0^\pi = 4$

$-2 \int_{0}^{\pi}{\sin^3(x)}dx = -2 \int_{0}^{\pi}{\sin^2(x) \sin(x)}dx= -\sin^2(x)\cos(x)-\int_{0}^{\pi}{\sin(2x)-\cos(x)}dx= -\sin^2(x)\cos(x)- (\sin(2x)-\sin(x) -\int_{0}^{\pi}{-\frac{1}{2}\cos(2x)(-\sin(x)dx}=...$

I've made a few steps more, but it doesn't simplify or repeat...

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As you did, write $\sin 2x=2\sin x\cos x$. Then let $u=\cos x$.

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As calculus, especially Integration of sum is much simpler than that of Product, why don't we use Werner Formulas

$$2\cos x\sin2x=\sin(2x+x)-\sin(2x-x)$$ and then $$\int\sin mx\ dx=-\frac{\cos mx}m+K$$

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  • $\begingroup$ @fear.xD, how about this? $\endgroup$ – lab bhattacharjee Feb 10 '14 at 13:10

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