Question Number 99173 by bemath last updated on 19/Jun/20
Answered by MJS last updated on 19/Jun/20
![this is very complicated ∫(√((1+cot x)/(csc x +cot x)))dx=∫(√((sin x +cos x)/(1+cos x)))dx= [t=tan (x/2) → dx=2cos^2 (x/2) dt] =(√2)∫((√(−t^2 +2t+1))/(t^2 +1))dt= [u=((√2)/2)(t−1) → dt=(√2)du] =(√2)∫((√(1−u^2 ))/(u^2 +(√2)u+1))du= [v=cos^(−1) u → du=−(√(1−u^2 ))dv] =−(√2)∫((sin^2 v)/(cos^2 v +(√2)cos v +1))dv= [w=tan (v/2) → dv=2cos^2 (v/2) dw] =−8(1+(√2))∫(w^2 /((w^2 +1)(w^4 +3+2(√2))))dw 2(√2)∫(dw/(w^2 +1))+(√(−2+2(√2)))(∫((w−(√(7+5(√2))))/(w^2 −(√(2+2(√2)))w+1+(√2)))dw−∫((w+(√(7+5(√2))))/(w^2 +(√(2+2(√2)))w+1+(√2)))dw) now just use formula](Q99199.png)
$$\mathrm{this}\:\mathrm{is}\:\mathrm{very}\:\mathrm{complicated} \\ $$$$\int\sqrt{\frac{\mathrm{1}+\mathrm{cot}\:{x}}{\mathrm{csc}\:{x}\:+\mathrm{cot}\:{x}}}{dx}=\int\sqrt{\frac{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:{dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{\sqrt{−{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({t}−\mathrm{1}\right)\:\rightarrow\:{dt}=\sqrt{\mathrm{2}}{du}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}{u}+\mathrm{1}}{du}= \\ $$$$\:\:\:\:\:\left[{v}=\mathrm{cos}^{−\mathrm{1}} \:{u}\:\rightarrow\:{du}=−\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }{dv}\right] \\ $$$$=−\sqrt{\mathrm{2}}\int\frac{\mathrm{sin}^{\mathrm{2}} \:{v}}{\mathrm{cos}^{\mathrm{2}} \:{v}\:+\sqrt{\mathrm{2}}\mathrm{cos}\:{v}\:+\mathrm{1}}{dv}= \\ $$$$\:\:\:\:\:\left[{w}=\mathrm{tan}\:\frac{{v}}{\mathrm{2}}\:\rightarrow\:{dv}=\mathrm{2cos}^{\mathrm{2}} \:\frac{{v}}{\mathrm{2}}\:{dw}\right] \\ $$$$=−\mathrm{8}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\int\frac{{w}^{\mathrm{2}} }{\left({w}^{\mathrm{2}} +\mathrm{1}\right)\left({w}^{\mathrm{4}} +\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{dw} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\int\frac{{dw}}{{w}^{\mathrm{2}} +\mathrm{1}}+\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}\left(\int\frac{{w}−\sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}}{{w}^{\mathrm{2}} −\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{w}+\mathrm{1}+\sqrt{\mathrm{2}}}{dw}−\int\frac{{w}+\sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}}{{w}^{\mathrm{2}} +\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{w}+\mathrm{1}+\sqrt{\mathrm{2}}}{dw}\right) \\ $$$$\mathrm{now}\:\mathrm{just}\:\mathrm{use}\:\mathrm{formula} \\ $$
Commented by bemath last updated on 19/Jun/20
$$\mathrm{yes}..\mathrm{sir}\:\mathrm{very}\:\mathrm{hard}... \\ $$