Question-211886

Question Number 211886 by Spillover last updated on 23/Sep/24

Answered by Ghisom last updated on 23/Sep/24

∫(√(cos^2 x +cos x)) dx=∫(√(2cos x)) cos (x/2) dx= [t=(((√2)sin (x/2))/( (√(cos x)))) ⇔ x=2arctan (t/( (√(t^2 +2)))) → dx=((√(2cos^3 x))/(cos (x/2)))dt] =2∫(dt/((t^2 +1)^2 ))=(t/(t^2 +1))+arctan t = =(√(2cos x)) sin (x/2) +arctan (((√2) sin (x/2))/( (√(cos x)))) +C

$$\int\sqrt{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{cos}\:{x}}\:{dx}=\int\sqrt{\mathrm{2cos}\:{x}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{cos}\:{x}}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{2cos}^{\mathrm{3}} \:{x}}}{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{arctan}\:{t}\:= \\ $$$$=\sqrt{\mathrm{2cos}\:{x}}\:\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:+\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{cos}\:{x}}}\:+{C} \\ $$

Answered by Spillover last updated on 23/Sep/24

(√(cos^2 x+cos x)) dx Let cos x=sin^2 t dx=((−2sin tcos t)/(sin x))dt=−((2sin tcos t)/( (√(1−cos^2 t))))dt ((2sin tcos t)/( (√(1−cos^2 t))))=−((2sin tcos t)/( (√(1−sin^4 t))))dt=−((2sin tcos t)/( (√((1−sin^2 t)(1+psin^2 t)))))dt dx=−((2sin tcos t)/( (√((1−sin^2 t)(1+sin^2 t)))))dt ∫(√(cos^2 x+cos x)) dx =∫(√(sin^4 t+sin^2 t)) dx =∫((√(sin^4 t+sin^2 t)) ×−((2sin tcos t)/( (√((1−sin^2 t)(1+sin^2 t)))))dt) −2∫sin^2 tdt=−2(((−sin tcos t+t)/2)) sin tcos t−t (√(cos x(1−cos x))) −sin^(−1) (√(cos x)) +C

$$\sqrt{\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}}\:{dx} \\ $$$${Let}\:\:\:\:\:\mathrm{cos}\:{x}=\mathrm{sin}\:^{\mathrm{2}} {t}\:\:\:\:\:\:{dx}=\frac{−\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\mathrm{sin}\:{x}}{dt}=−\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {t}}}{dt} \\ $$$$\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {t}}}=−\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{4}} {t}}}{dt}=−\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}\right)\left(\mathrm{1}+{p}\mathrm{sin}\:^{\mathrm{2}} {t}\right)}}{dt} \\ $$$${dx}=−\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}\right)\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {t}\right)}}{dt} \\ $$$$\int\sqrt{\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}}\:{dx}\:\:=\int\sqrt{\mathrm{sin}\:^{\mathrm{4}} {t}+\mathrm{sin}\:^{\mathrm{2}} {t}}\:{dx} \\ $$$$=\int\left(\sqrt{\mathrm{sin}\:^{\mathrm{4}} {t}+\mathrm{sin}\:^{\mathrm{2}} {t}}\:×−\frac{\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\:\sqrt{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}\right)\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {t}\right)}}{dt}\right) \\ $$$$−\mathrm{2}\int\mathrm{sin}\:^{\mathrm{2}} {tdt}=−\mathrm{2}\left(\frac{−\mathrm{sin}\:{t}\mathrm{cos}\:{t}+{t}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:{t}\mathrm{cos}\:{t}−{t} \\ $$$$\sqrt{\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}\:−\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{cos}\:{x}}\:+{C} \\ $$$$ \\ $$

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