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Question Number 121611 by benjo_mathlover last updated on 10/Nov/20

$$\underset{0}{\stackrel{2}{\int }}\sin \left(2\pi \mathrm{x}\right)\cos \left(5\pi \mathrm{x}\right)\mathrm{dx}?$$

Commented by mr W last updated on 10/Nov/20

$${\int }_0^2\sin \left(2\pi x\right)\cos \left(5\pi x\right)dx$$$$=\frac{1}{2}{\int }_0^2[\sin \left(7\pi x\right)-\sin \left(3\pi x\right)]dx$$$$=\frac{1}{2}{[-\frac{1}{7\pi }\cos \left(7\pi x\right)+\frac{1}{3\pi }\cos \left(3\pi x\right)]}_0^2$$$$=\frac{1}{2}[-\frac{1}{7\pi }(1-1)+\frac{1}{3\pi }(1-1)]$$$$=0$$

Commented by MJS_new last updated on 10/Nov/20

$$\sin \left(2\pi (2-x)\right)=-\sin \left(2\pi x\right)$$$$\cos \left(5\pi (2-x)\right)=\cos \left(5\pi x\right)$$$$\Rightarrow \underset{0}{\stackrel{2}{\int }}...=0$$

Answered by MJS_new last updated on 10/Nov/20

$$=0$$

Commented by benjo_mathlover last updated on 10/Nov/20

$$\mathrm{super}\mathrm{fastest}\mathrm{prof}$$

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