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Question Number 220184 by MathematicalUser2357 last updated on 07/May/25

∫_0 ^(π/(12)) (√((sec^4 α+5sec^5 αsin α)/((2−sec^2 α)(125tan^3 α+25tan^2 α+5tan α+1))))dα

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \sqrt{\frac{\mathrm{sec}^{\mathrm{4}} \alpha+\mathrm{5sec}^{\mathrm{5}} \alpha\mathrm{sin}\:\alpha}{\left(\mathrm{2}−\mathrm{sec}^{\mathrm{2}} \alpha\right)\left(\mathrm{125tan}^{\mathrm{3}} \alpha+\mathrm{25tan}^{\mathrm{2}} \alpha+\mathrm{5tan}\:\alpha+\mathrm{1}\right)}}{d}\alpha \\ $$

Answered by MathematicalUser2357 last updated on 07/May/25

Let x=tan α then dx=sec^2 αdα=(1−x^2 )dx Notice that sec^5 αsin α=sec^4 αtan α =∫_0 ^(2−(√3)) (√(((1−x^2 )^3 (1+x))/((2−(1−x^2 )^2 )(125x^3 +25x^2 +5x+1))))dx =∫_0 ^(2−(√3)) (√(((1−x^2 )^3 (1+x))/((2−(1−x^2 )^2 )(25x^2 +1)(5x+1))))dx =∫_0 ^(2−(√3)) (1−2x−7x^2 −4x^3 +((599x^4 )/2)+…)dx≈0.22884?

$$\mathrm{Let}\:{x}=\mathrm{tan}\:\alpha\:\mathrm{then}\:{dx}=\mathrm{sec}^{\mathrm{2}} \alpha{d}\alpha=\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$$\mathrm{Notice}\:\mathrm{that}\:\mathrm{sec}^{\mathrm{5}} \alpha\mathrm{sin}\:\alpha=\mathrm{sec}^{\mathrm{4}} \alpha\mathrm{tan}\:\alpha \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \sqrt{\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} \left(\mathrm{1}+{x}\right)}{\left(\mathrm{2}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \right)\left(\mathrm{125}{x}^{\mathrm{3}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\right)}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \sqrt{\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} \left(\mathrm{1}+{x}\right)}{\left(\mathrm{2}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \right)\left(\mathrm{25}{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{5}{x}+\mathrm{1}\right)}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \left(\mathrm{1}−\mathrm{2}{x}−\mathrm{7}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{3}} +\frac{\mathrm{599}{x}^{\mathrm{4}} }{\mathrm{2}}+\ldots\right){dx}\approx\mathrm{0}.\mathrm{22884}? \\ $$

Answered by Ghisom last updated on 08/May/25

∫(√((...)/(...)))dα= =∫(dα/( (√((1−2sin^2 α)(1+24sin^2 α)))))= [t=tan α] =∫(dt/( (√((1−t^2 )(1+25t^2 )))))= [u=arcsin t] =∫(du/( (√(1+25sin^2 u))))= =F (u∣−25) = =F (arcsin tan α ∣−25) +C

$$\int\sqrt{\frac{...}{...}}{d}\alpha= \\ $$$$=\int\frac{{d}\alpha}{\:\sqrt{\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}+\mathrm{24sin}^{\mathrm{2}} \:\alpha\right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\alpha\right] \\ $$$$=\int\frac{{dt}}{\:\sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{25}{t}^{\mathrm{2}} \right)}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsin}\:{t}\right] \\ $$$$=\int\frac{{du}}{\:\sqrt{\mathrm{1}+\mathrm{25sin}^{\mathrm{2}} \:{u}}}= \\ $$$$={F}\:\left({u}\mid−\mathrm{25}\right)\:= \\ $$$$={F}\:\left(\mathrm{arcsin}\:\mathrm{tan}\:\alpha\:\mid−\mathrm{25}\right)\:+{C} \\ $$

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