dx-csc-x-sec-x-

Question Number 132192 by benjo_mathlover last updated on 12/Feb/21

$$\:\int\:\frac{\mathrm{dx}}{\mathrm{csc}\:\mathrm{x}\:+\:\mathrm{sec}\:\mathrm{x}} \\ $$

Answered by Dwaipayan Shikari last updated on 12/Feb/21

∫((sinx cosx)/(sinx+cosx))dx =(1/2)∫sinx+cosx−(1/(sinx+cosx))dx =(1/2)(sinx−cosx)−∫(dt/(1−t^2 +2t)) t=tan(x/2) =(1/2)(sinx−cosx)+∫(1/((t−1)^2 −2))dt =(1/2)(sinx−cosx)+(1/(2(√2)))log(((t−1−(√2))/(t−1+(√2))))+C =(1/2)(sinx−cosx)+(1/(2(√2)))log(((tan(x/2)−1−(√2))/(tan(x/2)−1+(√2))))+C

$$\int\frac{{sinx}\:{cosx}}{{sinx}+{cosx}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{sinx}+{cosx}−\frac{\mathrm{1}}{{sinx}+{cosx}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}−{cosx}\right)−\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}}\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}−{cosx}\right)+\int\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}−{cosx}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}−{cosx}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{2}}}{{tan}\frac{{x}}{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}}}\right)+{C} \\ $$

Answered by MJS_new last updated on 13/Feb/21

weird but possible ∫(dx/(csc x +sec x))=∫((sin x cos x)/(sin x +cos x))dx= =((√2)/4)∫((sin 2x)/(sin ((4x+π)/4)))dx= [t=tan^2 ((4x+π)/8) → dx=((cos^2 ((4x+π)/8))/(tan ((4x+π)/8)))dt] =−((√2)/8)∫((t^2 −6t+1)/(t(t+1)^2 ))dt=(√2)∫(dt/((t+1)^2 ))−((√2)/8)∫(dt/t)= =−((√2)/(t+1))−((√2)/8)ln t = =−((√2)/4)ln ∣tan ((4x+π)/8)∣ −((√2)/2)cos ((4x+π)/4) +C

$$\mathrm{weird}\:\mathrm{but}\:\mathrm{possible} \\ $$$$\int\frac{{dx}}{\mathrm{csc}\:{x}\:+\mathrm{sec}\:{x}}=\int\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\frac{\mathrm{4}{x}+\pi}{\mathrm{4}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}^{\mathrm{2}} \:\frac{\mathrm{4}{x}+\pi}{\mathrm{8}}\:\rightarrow\:{dx}=\frac{\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{4}{x}+\pi}{\mathrm{8}}}{\mathrm{tan}\:\frac{\mathrm{4}{x}+\pi}{\mathrm{8}}}{dt}\right] \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}=\sqrt{\mathrm{2}}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{{dt}}{{t}}= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{{t}+\mathrm{1}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:{t}\:= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{tan}\:\frac{\mathrm{4}{x}+\pi}{\mathrm{8}}\mid\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{4}{x}+\pi}{\mathrm{4}}\:+{C} \\ $$

Answered by Ñï= last updated on 14/Feb/21

=∫((sinxcosx)/(sinx+cosx))dx =(1/2)∫((1+2sinxcosx−1)/(sinx+cosx))dx =(1/2)∫(sinx+cosx−(1/( (√2)))csc(x+(π/4)))dx =(1/2)(sinx−cosx)−(1/(2(√2)))ln∣tan((x/2)+(π/8))∣+C

$$=\int\frac{{sinxcosx}}{{sinx}+{cosx}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{2}{sinxcosx}−\mathrm{1}}{{sinx}+{cosx}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({sinx}+{cosx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{csc}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}−{cosx}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid+{C} \\ $$

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