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Question Number 92420 by I want to learn more last updated on 06/May/20

Commented by mathmax by abdo last updated on 07/May/20

I =∫ (dx/(cos^6 x +sin^6 x)) ⇒ I =∫ (dx/((cos^2 x)^3 +(sin^2 x)^3 )) =∫ (dx/((cos^2 x+sin^2 x)^3 −3cos^2 x sin^2 x(cos^2 x+sin^2 x))) =∫ (dx/(1−3cos^2 x sin^2 x)) =∫ (dx/(1−3((1/2)sin(2x))^2 )) =∫ (dx/(1−3×(1/4)sin^2 (2x))) =4 ∫ (dx/(4−3sin^2 (2x))) =4 ∫ (dx/(4−3×((1−cos(4x))/2))) =8 ∫ (dx/(8−3 +3cos(4x))) =8 ∫ (dx/(5+3cos(4x))) =_(4x=t) 2 ∫ (dt/(5+3cost)) =_(tan((t/2))=u) 2 ∫ (1/(5+3 ×((1−u^2 )/(1+u^2 ))))×((2du)/(1+u^2 )) =4 ∫ (du/(5+5u^2 +3−3u^2 )) =4 ∫ (du/(8+2u^2 )) =2 ∫ (du/(4+u^2 )) =_(u =4α) 2 ∫ ((4dα)/(4+4α^2 )) = 2 ∫ (dα/(1+α^2 )) =2arctan(α)+C =2arctan((u/4)) +C =2 arctan((1/4)tan((t/2))) +C =2arctan((1/4)tan(2x)) +C

$${I}\:=\int\:\:\frac{{dx}}{{cos}^{\mathrm{6}} {x}\:+{sin}^{\mathrm{6}} {x}}\:\Rightarrow\:{I}\:=\int\:\:\:\frac{{dx}}{\left({cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \:+\left({sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} } \\ $$$$=\int\:\:\frac{{dx}}{\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)} \\ $$$$=\int\:\:\frac{{dx}}{\mathrm{1}−\mathrm{3}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}}\:=\int\:\:\frac{{dx}}{\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} }\:=\int\:\:\frac{{dx}}{\mathrm{1}−\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dx}}{\mathrm{4}−\mathrm{3}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}\:=\mathrm{4}\:\int\:\:\:\frac{{dx}}{\mathrm{4}−\mathrm{3}×\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}}\: \\ $$$$=\mathrm{8}\:\int\:\:\:\frac{{dx}}{\mathrm{8}−\mathrm{3}\:+\mathrm{3}{cos}\left(\mathrm{4}{x}\right)}\:=\mathrm{8}\:\int\:\:\frac{{dx}}{\mathrm{5}+\mathrm{3}{cos}\left(\mathrm{4}{x}\right)}\:=_{\mathrm{4}{x}={t}} \:\mathrm{2}\:\int\:\:\frac{{dt}}{\mathrm{5}+\mathrm{3}{cost}} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\mathrm{2}\:\int\:\:\frac{\mathrm{1}}{\mathrm{5}+\mathrm{3}\:×\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}×\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int\:\:\:\frac{{du}}{\mathrm{5}+\mathrm{5}{u}^{\mathrm{2}} \:+\mathrm{3}−\mathrm{3}{u}^{\mathrm{2}} }\:=\mathrm{4}\:\int\:\:\:\frac{{du}}{\mathrm{8}+\mathrm{2}{u}^{\mathrm{2}} }\:=\mathrm{2}\:\int\:\:\frac{{du}}{\mathrm{4}+{u}^{\mathrm{2}} } \\ $$$$=_{{u}\:=\mathrm{4}\alpha} \:\:\:\mathrm{2}\:\int\:\:\frac{\mathrm{4}{d}\alpha}{\mathrm{4}+\mathrm{4}\alpha^{\mathrm{2}} }\:=\:\mathrm{2}\:\int\:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\mathrm{2}{arctan}\left(\alpha\right)+{C} \\ $$$$=\mathrm{2}{arctan}\left(\frac{{u}}{\mathrm{4}}\right)\:+{C}\:=\mathrm{2}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{4}}{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)\:+{C} \\ $$$$=\mathrm{2}{arctan}\left(\frac{\mathrm{1}}{\mathrm{4}}{tan}\left(\mathrm{2}{x}\right)\right)\:+{C} \\ $$

Commented by I want to learn more last updated on 12/May/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by niroj last updated on 06/May/20

∫ (( 1)/(cos^6 x+sin^6 x))dx = ∫ (( 1)/((sin^2 x)^3 +(cos^2 x)^3 ))dx = ∫ (( dx)/((sin^2 x+cos^2 x)(sin^4 x−sin^2 x cos^2 x+cos^4 x))) = ∫ (1/((sin^2 x+cos^2 x)^2 −2sin^2 x cos^2 x−sin^2 x cos^2 x))dx = ∫ (1/(1−3sin^2 xcos^2 x))dx = ∫ ((sec^2 x)/(sec^2 x−3tan^2 x))dx = ∫ ((sec^2 xdx)/(sec^2 x−tan^2 x−2tan^2 x)) = ∫ (( sec^2 x dx)/(1−2tan^2 x)) Put , tan x= t sec^2 xdx= dt ∫ (1/(1−2t^2 ))dt = (1/2)∫ (( 1)/((1/(2 )) −t^2 ))dt = (1/2)∫ (( 1)/(((√(1/2)) )^2 −(t)^2 ))dt = (1/2) .(1/(2(√(1/2)))) log (( (√(1/2)) +t)/( (√(1/2))−t)) +C = (1/2)×(1/(2×(1/(√2))))log (((1/(√2)) +t)/((1/(√2))−t))+C = ((√2)/4) log ((1+ t(√2))/(1−t(√2))) +C = ((√2)/4)×((√2)/(√2)) log ((1+tan x(√2))/(1−tan x (√2))) +C = (1/(2(√2))) log ((1+ tan x (√2))/(1−tan x (√2))) +C //.

$$\:\:\:\:\int\:\frac{\:\:\mathrm{1}}{\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\:\:\mathrm{1}}{\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} +\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$$$=\:\int\:\frac{\:\:\mathrm{dx}}{\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{sin}^{\mathrm{4}} \mathrm{x}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{4}} \mathrm{x}\right)} \\ $$$$=\:\int\:\frac{\mathrm{1}}{\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} −\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{sec}^{\mathrm{2}} \mathrm{x}−\mathrm{3tan}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\:=\:\int\:\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\mathrm{sec}^{\mathrm{2}} \mathrm{x}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}−\mathrm{2tan}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:=\:\int\:\frac{\:\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\mathrm{1}−\mathrm{2tan}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\mathrm{Put}\:,\:\:\mathrm{tan}\:\mathrm{x}=\:\mathrm{t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}=\:\mathrm{dt} \\ $$$$\:\:\:\:\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\:\:\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}\:}\:−\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\:\:\mathrm{1}}{\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\:\:\right)^{\mathrm{2}} −\left(\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:.\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}\:\mathrm{log}\:\frac{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\:+\mathrm{t}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}−\mathrm{t}}\:+\mathrm{C} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\mathrm{log}\:\:\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\mathrm{t}}{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{t}}+\mathrm{C} \\ $$$$\:\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{log}\:\:\frac{\mathrm{1}+\:\mathrm{t}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}\sqrt{\mathrm{2}}}\:+\mathrm{C} \\ $$$$\:\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}}\:\mathrm{log}\:\:\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{x}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{2}}}\:+\mathrm{C} \\ $$$$\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{log}\:\:\frac{\mathrm{1}+\:\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{2}}}\:+\mathrm{C}\://. \\ $$

Commented by I want to learn more last updated on 06/May/20

Thanks sir. I appreciate

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

Commented by niroj last updated on 06/May/20

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Answered by niroj last updated on 07/May/20

Such type also you can Apply: ∫ (1/(cos^6 x+sin^6 x))dx dived numerator & denominator by cos^6 x. = ∫ ((sec^6 x dx)/(1+tan^6 x)) = ∫ (( sec^4 x. sec^2 x dx)/((1)^3 +(tan^2 x)^3 )) = ∫(((sec^2 x)^2 sec^2 xdx)/((1+tan^2 x)(1−tan^2 x+tan^4 x))) = ∫(((1+tan^2 x)^2 sec^2 xdx)/((1+tan^2 x)(1−tan^2 x+tan^4 x))) Put , tan x=t sec^2 xdx=dt ∫ (((1+t^2 )^2 dt)/((1+t^2 )(t^4 −t^2 +1))) = ∫ (((1+t^2 )dt)/(t^4 −t^2 +1)) = ∫ ((t^2 (1+(1/t^2 )))/(t^2 (t^2 −1+(1/t^2 ))))dt = ∫ (((1+(1/t^2 ))dt)/(t^2 −1+(1/t^2 )))= ∫ (((1+(1/t^2 ))dt)/((t−(1/t))^2 +2−1)) Put, (t−(1/t))=m (1+(1/t^2 ))dt=dm ∫ (( 1)/(m^2 +1))dm= tan^(−1) m +C = tan^(−1) (t−(1/t))+C = tan^(−1) (tan x −(1/(tan x)))+C = tan^(−1) (tan x−cot x) +C //.

$$\mathrm{Such}\:\mathrm{type}\:\mathrm{also}\:\mathrm{you}\:\mathrm{can}\:\mathrm{Apply}: \\ $$$$\:\:\:\int\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}}\mathrm{dx} \\ $$$$\:\:\mathrm{dived}\:\mathrm{numerator}\:\&\:\mathrm{denominator}\:\mathrm{by}\:\mathrm{cos}^{\mathrm{6}} \mathrm{x}. \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{sec}^{\mathrm{6}} \mathrm{x}\:\mathrm{dx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{6}} \mathrm{x}} \\ $$$$\:\:=\:\int\:\frac{\:\:\mathrm{sec}^{\mathrm{4}} \mathrm{x}.\:\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\left(\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} } \\ $$$$\:=\:\int\frac{\left(\mathrm{sec}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} \:\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{tan}^{\mathrm{4}} \mathrm{x}\right)} \\ $$$$=\:\int\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} \mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{tan}^{\mathrm{4}} \mathrm{x}\right)} \\ $$$$\:\:\mathrm{Put}\:,\:\:\mathrm{tan}\:\mathrm{x}=\mathrm{t} \\ $$$$\:\:\:\:\:\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}=\mathrm{dt} \\ $$$$\:\:\:\int\:\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} −\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:=\:\int\:\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} −\mathrm{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\:\int\:\frac{\mathrm{t}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}\mathrm{dt} \\ $$$$\:=\:\int\:\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}=\:\int\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}−\mathrm{1}} \\ $$$$\:\mathrm{Put},\:\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)=\mathrm{m} \\ $$$$\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}=\mathrm{dm} \\ $$$$\:\:\:\:\int\:\frac{\:\mathrm{1}}{\mathrm{m}^{\mathrm{2}} +\mathrm{1}}\mathrm{dm}=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{m}\:+\mathrm{C} \\ $$$$\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)+\mathrm{C} \\ $$$$\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{x}\:−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{x}}\right)+\mathrm{C} \\ $$$$\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}\right)\:+\mathrm{C}\://. \\ $$$$ \\ $$

Commented by I want to learn more last updated on 07/May/20

Thanks sir, i appreciate your time

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

Commented by niroj last updated on 07/May/20

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