Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 152841 by 0731619 last updated on 02/Sep/21

Answered by bobhans last updated on 02/Sep/21

(2)(1/4)∫ sin^2 (2x) dx =        (1/4)∫ ((1−cos (4x))/2) dx =        (1/4)((x/2)−(1/8)sin (4x))+ c =       (x/8)−((sin (4x))/(32)) + c

$$\left(\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{4}}\int\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{dx}\:= \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{4}{x}\right)}{\mathrm{2}}\:{dx}\:= \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\left(\mathrm{4}{x}\right)\right)+\:{c}\:= \\ $$$$\:\:\:\:\:\frac{{x}}{\mathrm{8}}−\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{32}}\:+\:{c} \\ $$

Answered by bobhans last updated on 02/Sep/21

(1) ∫ ((2sin 2θ)/(cos (2θ )(√(1+3cos^2 (2θ))))) dθ =     ∫ ((−d(cos (2θ)))/(cos (2θ)(√(1+3cos^2 (2θ))))) =     ∫ ((−du)/(u (√(1+3u^2 )))) =∫ ((−1)/u) du +∫ (1/( (√(1+3u^2 )))) du  = −ln ∣u∣ +∫ (du/( (√(1+(u(√3))^2 ))))   =−ln ∣cos 2(θ)∣ +(1/( (√3))) ln ∣(√(1+3cos^2 (2θ))) (√3) cos (2θ)∣ + c  =ln ∣sec (2θ)∣+(1/( (√3))) ln ∣cos (2θ)(√(3+9cos^2 (2θ))) ∣+c

$$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}\theta}{\mathrm{cos}\:\left(\mathrm{2}\theta\:\right)\sqrt{\mathrm{1}+\mathrm{3cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)}}\:{d}\theta\:= \\ $$$$\:\:\:\int\:\frac{−{d}\left(\mathrm{cos}\:\left(\mathrm{2}\theta\right)\right)}{\mathrm{cos}\:\left(\mathrm{2}\theta\right)\sqrt{\mathrm{1}+\mathrm{3cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)}}\:= \\ $$$$\:\:\:\int\:\frac{−{du}}{{u}\:\sqrt{\mathrm{1}+\mathrm{3}{u}^{\mathrm{2}} }}\:=\int\:\frac{−\mathrm{1}}{{u}}\:{du}\:+\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{3}{u}^{\mathrm{2}} }}\:{du} \\ $$$$=\:−\mathrm{ln}\:\mid{u}\mid\:+\int\:\frac{{du}}{\:\sqrt{\mathrm{1}+\left({u}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}\: \\ $$$$=−\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2}\left(\theta\right)\mid\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{ln}\:\mid\sqrt{\mathrm{1}+\mathrm{3cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)}\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\mid\:+\:{c} \\ $$$$=\mathrm{ln}\:\mid\mathrm{sec}\:\left(\mathrm{2}\theta\right)\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{ln}\:\mid\mathrm{cos}\:\left(\mathrm{2}\theta\right)\sqrt{\mathrm{3}+\mathrm{9cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)}\:\mid+{c} \\ $$

Commented by peter frank last updated on 03/Sep/21

good

$$\mathrm{good} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com