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Question Number 170831 by venom1 last updated on 01/Jun/22

Answered by LEKOUMA last updated on 01/Jun/22

1) ∫4sin 8xdx=4∫sin 8xdx=4×−(1/8)cos 8x+c=−(1/2)cos 8x+c 2) ∫xcos (12x^2 )dx let u=12x^2 ⇒ du=24xdx ⇒ dx=(du/(24x)) ∫(1/(24))cos( u)du=(1/(24))∫cos (u)du=(1/(24))sin( u)+c ∫xcos (12x^2 )dx=(1/(24))sin (12x^2 )+c 3) ∫(4x−6)sin (x^2 −3x+4)dx ∫2(2x−3)sin (x^2 −3x+4)dx 2∫(2x−3)sin (x^2 −3x+4)dx let u=x^2 −3x+4 ⇒ du=(2x−3)dx ⇒ (du/((2x−3)))=dx 2∫sin udu=−2cos u+c=−2cos (x^2 −3x+4)+c

$$\left.\mathrm{1}\right)\:\int\mathrm{4sin}\:\mathrm{8}{xdx}=\mathrm{4}\int\mathrm{sin}\:\mathrm{8}{xdx}=\mathrm{4}×−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{8}{x}+{c}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{8}{x}+{c} \\ $$$$\left.\mathrm{2}\right)\:\int{x}\mathrm{cos}\:\left(\mathrm{12}{x}^{\mathrm{2}} \right){dx} \\ $$$${let}\:{u}=\mathrm{12}{x}^{\mathrm{2}} \:\Rightarrow\:{du}=\mathrm{24}{xdx}\:\Rightarrow\:{dx}=\frac{{du}}{\mathrm{24}{x}} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{24}}\mathrm{cos}\left(\:{u}\right){du}=\frac{\mathrm{1}}{\mathrm{24}}\int\mathrm{cos}\:\left({u}\right){du}=\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}\left(\:{u}\right)+{c} \\ $$$$\int{x}\mathrm{cos}\:\left(\mathrm{12}{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}\:\left(\mathrm{12}{x}^{\mathrm{2}} \right)+{c} \\ $$$$\left.\mathrm{3}\right)\:\int\left(\mathrm{4}{x}−\mathrm{6}\right)\mathrm{sin}\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right){dx} \\ $$$$\int\mathrm{2}\left(\mathrm{2}{x}−\mathrm{3}\right)\mathrm{sin}\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right){dx} \\ $$$$\mathrm{2}\int\left(\mathrm{2}{x}−\mathrm{3}\right)\mathrm{sin}\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right){dx} \\ $$$${let}\:{u}={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\:\Rightarrow\:{du}=\left(\mathrm{2}{x}−\mathrm{3}\right){dx}\:\Rightarrow\:\frac{{du}}{\left(\mathrm{2}{x}−\mathrm{3}\right)}={dx} \\ $$$$\mathrm{2}\int\mathrm{sin}\:{udu}=−\mathrm{2cos}\:{u}+{c}=−\mathrm{2cos}\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)+{c} \\ $$

Answered by LEKOUMA last updated on 01/Jun/22

6)∫sin tcos (cos t)dt let u=cos t ⇒ du=−sin tdt ⇒ (du/(−sin t))=dt ∫−cos udu=−∫cos udu=−sin u+c =−sin (cos t)+c

$$\left.\mathrm{6}\right)\int\mathrm{sin}\:{t}\mathrm{cos}\:\left(\mathrm{cos}\:{t}\right){dt} \\ $$$${let}\:{u}=\mathrm{cos}\:{t}\:\Rightarrow\:{du}=−\mathrm{sin}\:{tdt}\:\Rightarrow\:\frac{{du}}{−\mathrm{sin}\:{t}}={dt} \\ $$$$\int−\mathrm{cos}\:{udu}=−\int\mathrm{cos}\:{udu}=−\mathrm{sin}\:{u}+{c} \\ $$$$=−\mathrm{sin}\:\left(\mathrm{cos}\:{t}\right)+{c} \\ $$

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