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Question Number 83989 by Rio Michael last updated on 08/Mar/20

$$\mathrm{find}$$$$\mathrm{a}.\int \cos 3x\cos 5xdx$$$$\mathrm{b}.\int x\ln 2xdx$$

Commented by jagoll last updated on 08/Mar/20

$$\mathrm{a}.\int \frac{1}{2}(\cos 8\mathrm{x}+\cos 2\mathrm{x})\mathrm{dx}=$$$$\frac{1}{16}\sin 8\mathrm{x}+\frac{1}{4}\sin 2\mathrm{x}+\mathrm{c}$$

Commented by jagoll last updated on 08/Mar/20

$$\mathrm{b}.\mathrm{by}\mathrm{part}$$$$\int \mathrm{x}\ln \left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{x}}^2\ln \left(2\mathrm{x}\right)-\int \frac{1}{2}{\mathrm{x}}^2\left(\frac{1}{\mathrm{x}}\right)\mathrm{dx}$$$$=\frac{1}{2}{\mathrm{x}}^2\ln \left(2\mathrm{x}\right)-\frac{1}{2}\int \mathrm{x}\mathrm{dx}$$$$=\frac{1}{2}{\mathrm{x}}^2\ln \left(2\mathrm{x}\right)-\frac{1}{4}{\mathrm{x}}^2+\mathrm{c}$$$$=\frac{1}{4}{\mathrm{x}}^2(2\ln \left(2\mathrm{x}\right)-1)+\mathrm{c}$$$$=\frac{1}{4}{\mathrm{x}}^2\ln \left(\frac{{4\mathrm{x}}^2}{\mathrm{e}}\right)+\mathrm{c}$$

Commented by Rio Michael last updated on 08/Mar/20

$$thankssir$$

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