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Question Number 77002 by jagoll last updated on 02/Jan/20
calculate ∫ (dt/(cos (2t+a)cos (2t−a)))
$$ \\ $$$${calculate}\:\int\:\frac{{dt}}{\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)} \\ $$
Commented by turbo msup by abdo last updated on 03/Jan/20
we have cos(2t+a)cos(2t−a) =(1/2)(cos(4t)+cos(2a)) ⇒ I =2∫ (dt/(cos(4t)+cos(2a))) =_(4t=u) 2 ∫ (1/(cos(u)+cos(2a)))(du/4) =(1/2)∫ (du/(cos(u)+cos(2a))) =_(tan((u/2))=x) (1/2) ∫ (1/(((1−x^2 )/(1+x^2 )) +cos(2a)))((2dx)/(1+x^2 )) =∫ (dx/(1−x^2 +cos(2a)+cos(2a)x^2 )) =∫ (dx/((cos(2a)−1)x^2 +1+cos(2a))) =∫ (dx/(1+cos(2a)−(1−cos(2a)x^2 )) =(1/(1+cos(2a))) ∫ (dx/(1−((1−cos(2a))/(1+cos(2a)))x^2 )) =(1/(1+cos(2a))) ∫ (dx/(1−(tana x)^2 )) =_(x tana =z) (1/(1+cos(2a))) ∫ (dz/(tana(1−z^2 ))) =(1/(1+cos(2a))tana)) ∫ ((1/(1−z))+(1/(1+z)))dz =(1/(tana(1+cos(2a))))(1/2)ln∣((1+z)/(1−z))∣ +c =(1/(2tana(1+cos(2a))))ln∣((1+xtana)/(1−xtana))∣ +c =(1/(2tana(1+cos(2a)))ln∣((1+tanatan((u/2)))/(1−xtana tan((u/2))))∣ +C with u=4t
$${we}\:{have}\: \\ $$$${cos}\left(\mathrm{2}{t}+{a}\right){cos}\left(\mathrm{2}{t}−{a}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{4}{t}\right)+{cos}\left(\mathrm{2}{a}\right)\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\int\:\frac{{dt}}{{cos}\left(\mathrm{4}{t}\right)+{cos}\left(\mathrm{2}{a}\right)} \\ $$$$=_{\mathrm{4}{t}={u}} \:\:\:\mathrm{2}\:\int\:\:\frac{\mathrm{1}}{{cos}\left({u}\right)+{cos}\left(\mathrm{2}{a}\right)}\frac{{du}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{du}}{{cos}\left({u}\right)+{cos}\left(\mathrm{2}{a}\right)} \\ $$$$=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:+{cos}\left(\mathrm{2}{a}\right)}\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} \:+{cos}\left(\mathrm{2}{a}\right)+{cos}\left(\mathrm{2}{a}\right){x}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{{dx}}{\left({cos}\left(\mathrm{2}{a}\right)−\mathrm{1}\right){x}^{\mathrm{2}} \:+\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)} \\ $$$$=\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)−\left(\mathrm{1}−{cos}\left(\mathrm{2}{a}\right){x}^{\mathrm{2}} \right.} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)}\:\int\:\:\:\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}−{cos}\left(\mathrm{2}{a}\right)}{\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)}\:\int\:\:\frac{{dx}}{\mathrm{1}−\left({tana}\:{x}\right)^{\mathrm{2}} } \\ $$$$=_{{x}\:{tana}\:={z}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)}\:\int\:\:\frac{{dz}}{{tana}\left(\mathrm{1}−{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\left.\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)\right){tana}}\:\int\:\:\left(\frac{\mathrm{1}}{\mathrm{1}−{z}}+\frac{\mathrm{1}}{\mathrm{1}+{z}}\right){dz} \\ $$$$=\frac{\mathrm{1}}{{tana}\left(\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)\right)}\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{z}}{\mathrm{1}−{z}}\mid\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{tana}\left(\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)\right)}{ln}\mid\frac{\mathrm{1}+{xtana}}{\mathrm{1}−{xtana}}\mid\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{tana}\left(\mathrm{1}+{cos}\left(\mathrm{2}{a}\right)\right.}{ln}\mid\frac{\mathrm{1}+{tanatan}\left(\frac{{u}}{\mathrm{2}}\right)}{\mathrm{1}−{xtana}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)}\mid\:+{C} \\ $$$${with}\:{u}=\mathrm{4}{t} \\ $$
Commented by mathmax by abdo last updated on 03/Jan/20
⇒I =(1/(2tan(a)×(2cos^2 a)))ln∣((1+tana.tan(2t))/(1−tana.tan(2t)))∣ +C =(1/(4 ((sina)/(cosa))×cos^2 a))ln∣((1+tan(a)tan(2t))/(1−tana.tan(2t)))∣ +C =(1/(2sin(2a)))ln∣((1+tana.tan(2t))/(1−tana.tan(2t)))∣ +C.
$$\Rightarrow{I}\:=\frac{\mathrm{1}}{\mathrm{2}{tan}\left({a}\right)×\left(\mathrm{2}{cos}^{\mathrm{2}} {a}\right)}{ln}\mid\frac{\mathrm{1}+{tana}.{tan}\left(\mathrm{2}{t}\right)}{\mathrm{1}−{tana}.{tan}\left(\mathrm{2}{t}\right)}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\:\frac{{sina}}{{cosa}}×{cos}^{\mathrm{2}} {a}}{ln}\mid\frac{\mathrm{1}+{tan}\left({a}\right){tan}\left(\mathrm{2}{t}\right)}{\mathrm{1}−{tana}.{tan}\left(\mathrm{2}{t}\right)}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\mathrm{2}{a}\right)}{ln}\mid\frac{\mathrm{1}+{tana}.{tan}\left(\mathrm{2}{t}\right)}{\mathrm{1}−{tana}.{tan}\left(\mathrm{2}{t}\right)}\mid\:+{C}. \\ $$
Answered by john santu last updated on 02/Jan/20
by manipulate algebra =∫ ((sin[(2t+a)−(2t−a)])/(sin 2a×cos (2t+a)cos (2t−a)))dt = (1/(sin 2a))∫ ((sin (2t+a)cos (2t−a)−cos (2t+a)sin (2t−a))/(cos (2t+a)cos (2t−a))) dt =(1/(sin 2a)){∫tan (2t+a)dt−∫tan (2t−a)dt} = (1/(sin 2a)){−(1/2)ln∣sec (2t+a)∣+(1/2)ln∣sec (2t−a)∣}+c .■★
$${by}\:{manipulate}\:{algebra}\: \\ $$$$=\int\:\frac{{sin}\left[\left(\mathrm{2}{t}+{a}\right)−\left(\mathrm{2}{t}−{a}\right)\right]}{\mathrm{sin}\:\mathrm{2}{a}×\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{a}}\int\:\frac{\mathrm{sin}\:\left(\mathrm{2}{t}+{a}\right)\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)−\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)\mathrm{sin}\:\left(\mathrm{2}{t}−{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{a}}\left\{\int\mathrm{tan}\:\left(\mathrm{2}{t}+{a}\right){dt}−\int\mathrm{tan}\:\left(\mathrm{2}{t}−{a}\right){dt}\right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{a}}\left\{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{sec}\:\left(\mathrm{2}{t}+{a}\right)\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{sec}\:\left(\mathrm{2}{t}−{a}\right)\mid\right\}+{c}\:.\blacksquare\bigstar \\ $$$$ \\ $$
Commented by peter frank last updated on 02/Jan/20
good
$${good} \\ $$
Commented by jagoll last updated on 02/Jan/20
thanks you sir
$${thanks}\:{you}\:{sir} \\ $$
Commented by peter frank last updated on 02/Jan/20
please help Qn 76964
$${please}\:{help}\:{Qn}\:\mathrm{76964} \\ $$
Commented by john santu last updated on 02/Jan/20
i try sir
$${i}\:{try}\:{sir} \\ $$$$ \\ $$
Answered by MJS last updated on 02/Jan/20
∫(dt/(cos (2t+a) cos (2t−a)))= =2∫(dt/(cos 4t +cos 2a))= [u=tan 2t → dt=(du/(2(t^2 +1)))] =−(1/2)∫(du/(u^2 sin^2 a −cos^2 a ))= =(1/(4cos a))(∫(du/(usin a +cos a))−∫(du/(usin a −cos a)))= =(1/(4cos a sin a))(ln (usin a +cos a) −ln (usin a −cos a))= =(1/(2sin 2a))ln ((usin a +cos a)/(usin a −cos a)) = =(1/(2sin 2a))ln ∣((cos (2t−a))/(cos (2t+a)))∣ +C
$$\int\frac{{dt}}{\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)\:\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)}= \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{cos}\:\mathrm{4}{t}\:+\mathrm{cos}\:\mathrm{2}{a}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\mathrm{2}{t}\:\rightarrow\:{dt}=\frac{{du}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{a}\:−\mathrm{cos}^{\mathrm{2}} \:{a}\:}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4cos}\:{a}}\left(\int\frac{{du}}{{u}\mathrm{sin}\:{a}\:+\mathrm{cos}\:{a}}−\int\frac{{du}}{{u}\mathrm{sin}\:{a}\:−\mathrm{cos}\:{a}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4cos}\:{a}\:\mathrm{sin}\:{a}}\left(\mathrm{ln}\:\left({u}\mathrm{sin}\:{a}\:+\mathrm{cos}\:{a}\right)\:−\mathrm{ln}\:\left({u}\mathrm{sin}\:{a}\:−\mathrm{cos}\:{a}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2sin}\:\mathrm{2}{a}}\mathrm{ln}\:\frac{{u}\mathrm{sin}\:{a}\:+\mathrm{cos}\:{a}}{{u}\mathrm{sin}\:{a}\:−\mathrm{cos}\:{a}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2sin}\:\mathrm{2}{a}}\mathrm{ln}\:\mid\frac{\mathrm{cos}\:\left(\mathrm{2}{t}−{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{t}+{a}\right)}\mid\:+{C} \\ $$

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