Menu Close

Question-156739




Question Number 156739 by cortano last updated on 15/Oct/21
Commented by john_santu last updated on 15/Oct/21
 B=∫ (√((sin x+cos x)/(sin x−cos x))) dx     =∫ (√((sin (x+(π/4)))/(−cos (x+(π/4))))) dx    =∫ (√(−tan  (x+(π/4)))) dx    =∫(√(cot (x+((3π)/4)))) d(x+((3π)/4))   =∫ (dq/( (√(tan q))))    let y=(√(tan q)) ⇒q=tan^(−1) (y^2 )   dq=((2y)/(1+y^4 )) dy    B=∫ (1/y).((2y)/(1+y^4 )) dy=∫ ((2dy)/(1+y^4 ))   = ∫ (((y^2 +1)−(y^2 −1))/(1+y^4 )) dy   =∫ ((1+(1/y^2 ))/(y^2 +(1/y^2 )))dy−∫((1−y^2 )/(y^2 +(1/y^2 )))dy   =∫ ((d(y−(1/y)))/((y−(1/y))^2 +2))+∫ ((d(y+(1/y)))/(2−(y+(1/y))^2 ))   =(1/( (√2))) tan^(−1) ((1/( (√2)))(y−(1/y)))+(1/( (√2)))tanh^(−1) ((1/( (√2)))(y+(1/y)))+C   =(1/( (√2)))tan^(−1) ((1/( (√2)))((√(tan q))−(1/( (√(tan q))))))+(1/( (√2)))tanh^(−1) ((1/( (√2)))((√(tan q))+(1/( (√(tan q))))))+C   =(1/( (√2)))tan^(−1) ((1/( (√2)))((√(tan (x+((3π)/4))))−(1/( (√(tan (x+((3π)/4)))))))     +(1/( (√2)))tanh^(−1) ((1/( (√2)))((√(tan (x+((3π)/4))))+(1/( (√(tan (x+((3π)/4)))))))+C
$$\:{B}=\int\:\sqrt{\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}}\:{dx} \\ $$$$\:\:\:=\int\:\sqrt{\frac{\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}{−\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}}\:{dx} \\ $$$$\:\:=\int\:\sqrt{−\mathrm{tan}\:\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}\:{dx} \\ $$$$\:\:=\int\sqrt{\mathrm{cot}\:\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:{d}\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right) \\ $$$$\:=\int\:\frac{{dq}}{\:\sqrt{\mathrm{tan}\:{q}}}\: \\ $$$$\:{let}\:{y}=\sqrt{\mathrm{tan}\:{q}}\:\Rightarrow{q}=\mathrm{tan}^{−\mathrm{1}} \left({y}^{\mathrm{2}} \right) \\ $$$$\:{dq}=\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{4}} }\:{dy}\: \\ $$$$\:{B}=\int\:\frac{\mathrm{1}}{{y}}.\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{4}} }\:{dy}=\int\:\frac{\mathrm{2}{dy}}{\mathrm{1}+{y}^{\mathrm{4}} } \\ $$$$\:=\:\int\:\frac{\left({y}^{\mathrm{2}} +\mathrm{1}\right)−\left({y}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{1}+{y}^{\mathrm{4}} }\:{dy} \\ $$$$\:=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}{{y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}{dy}−\int\frac{\mathrm{1}−{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}{dy} \\ $$$$\:=\int\:\frac{{d}\left({y}−\frac{\mathrm{1}}{{y}}\right)}{\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\mathrm{2}}+\int\:\frac{{d}\left({y}+\frac{\mathrm{1}}{{y}}\right)}{\mathrm{2}−\left({y}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} } \\ $$$$\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({y}−\frac{\mathrm{1}}{{y}}\right)\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({y}+\frac{\mathrm{1}}{{y}}\right)\right)+{C} \\ $$$$\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{\mathrm{tan}\:{q}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:{q}}}\right)\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{\mathrm{tan}\:{q}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:{q}}}\right)\right)+{C} \\ $$$$\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{\mathrm{tan}\:\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}}\right)\right. \\ $$$$\:\:\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{\mathrm{tan}\:\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\left({x}+\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}}\right)+{C}\right. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *