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

find a. ∫cos 3x cos 5x dx b. ∫xln 2x dx

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

Commented by jagoll last updated on 08/Mar/20

a. ∫ (1/2)(cos 8x+cos 2x) dx = (1/(16)) sin 8x+(1/4)sin 2x + c

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

Commented by jagoll last updated on 08/Mar/20

b. by part ∫ x ln(2x) dx = (1/2)x^2 ln(2x)−∫ (1/2)x^2 ((1/x)) dx = (1/2)x^2 ln(2x)−(1/2)∫ x dx = (1/2)x^2 ln(2x) − (1/4)x^2 + c = (1/4)x^2 ( 2 ln(2x)−1) + c = (1/4)x^2 ln (((4x^2 )/e)) + c

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

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

thanks sir

$${thanks}\:{sir} \\ $$

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