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Question Number 93304 by Shakhzod last updated on 12/May/20
∫(√(tan (x)))dx Who can solve this problem?
$$\int\sqrt{\mathrm{tan}\:\left({x}\right)}{dx}\:{Who}\:{can}\:{solve}\:{this}\:{problem}? \\ $$
Commented by john santu last updated on 12/May/20
= −(1/( (√2) ))tanh^(−1) ((((√(tan x))+(√(cot x)))/( (√2)))) + (1/( (√2)))tan^(−1) ((((√(tan x))−(√(cot x)))/( (√2))))+c
$$=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right)\:+\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{c}\:\: \\ $$
Commented by i jagooll last updated on 12/May/20
cooll man ����
Commented by i jagooll last updated on 12/May/20
set u = (√(tan x)) , dx = ((2u du)/(u^4 +1)) ∫ (√(tan x)) dx = ∫ (2/(u^2 +(1/u^2 ))) du = ∫ ((1−(1/u))/((u+(1/u))^2 −2)) du + ∫ ((1+(1/u^2 ))/((u−(1/u))^2 +2)) du [ set u+(1/u) = tan q ], [ w=u−(1/u)] = ∫ (1/(q^2 −2)) dq + ∫ (1/(w^2 +2)) dw =−(1/( (√2))) tanh^(−1) ((q/( (√2)))) +(1/( (√2))) tan^(−1) ((w/( (√2)))) + c = −(1/( (√2))) tanh^(−1) ((((√(tan x))+(√(cot x)))/( (√2)))) + (1/( (√2))) tan^(−1) ((((√(tan x))−(√(cot x)))/( (√2)))) + c
$$\mathrm{set}\:\mathrm{u}\:=\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:,\:\mathrm{dx}\:=\:\frac{\mathrm{2u}\:\mathrm{du}}{\mathrm{u}^{\mathrm{4}} +\mathrm{1}} \\ $$$$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }}\:\mathrm{du} \\ $$$$=\:\int\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{u}}}{\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{2}} −\mathrm{2}}\:\mathrm{du}\:+\:\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }}{\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{2}} +\mathrm{2}}\:\mathrm{du} \\ $$$$\left[\:\mathrm{set}\:\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}}\:=\:\mathrm{tan}\:\mathrm{q}\:\right],\:\left[\:\mathrm{w}=\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\right] \\ $$$$=\:\int\:\frac{\mathrm{1}}{\mathrm{q}^{\mathrm{2}} −\mathrm{2}}\:\mathrm{dq}\:+\:\int\:\frac{\mathrm{1}}{\mathrm{w}^{\mathrm{2}} +\mathrm{2}}\:\mathrm{dw}\: \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{q}}{\:\sqrt{\mathrm{2}}}\right)\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{w}}{\:\sqrt{\mathrm{2}}}\right)\:+\:\mathrm{c} \\ $$$$=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right)\:+\:\mathrm{c}\: \\ $$
Commented by  M±th+et+s last updated on 12/May/20
at end pls check for typos and thank you
$${at}\:{end}\:{pls}\:{check}\:{for}\:{typos}\:{and}\:{thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 12/May/20
this integral is solved see the platform.
$${this}\:{integral}\:{is}\:{solved}\:{see}\:{the}\:{platform}. \\ $$
Answered by prakash jain last updated on 12/May/20
Commented by Shakhzod last updated on 12/May/20
Great. Thank you so much.
$${Great}.\:{Thank}\:{you}\:{so}\:{much}. \\ $$
Answered by  M±th+et+s last updated on 12/May/20
=(1/( (√2)))∫((2sin(x))/( (√(sin(2x))))) (1/( (√2)))∫((sin(x)+cos(x)+sin(x)−cos(x))/( (√(sin(2x))))).((cos(2x))/(cos(2x)))dx =(1/( (√2)))∫((1+(√(sin(2x))))/( (√(1−sin^2 (2x))))).((cos(2x))/( (√(sin(2x)))))dx+(1/( (√2)))∫((√(1−sin(2x)))/( (√(1−sin^2 (2x))))).((cos(2x))/( (√(sin(2x)))))dx =(1/( (√2)))∫(1/( (√(1−((√(sin(2x))))^2 )))).((cos(2x))/( (√(sin(2x)))))+(1/( (√2)))∫(1/( (√(1+((√(sin(2x))))^2 )))).((cos(2x))/( (√(sin(2x)))))dx =(1/( (√2)))sin^(−1) ((√(sin(2x))))+(1/( (√2)))sinh^(−1) ((√(sin(2x))))+c i can solve it with using another ways but i dont have a time now i will post them in anothe time
$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{2}{sin}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{sin}\left({x}\right)+{cos}\left({x}\right)+{sin}\left({x}\right)−{cos}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}.\frac{{cos}\left(\mathrm{2}{x}\right)}{{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}+\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}}.\frac{{cos}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\sqrt{\mathrm{1}−{sin}\left(\mathrm{2}{x}\right)}}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}}.\frac{{cos}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\sqrt{{sin}\left(\mathrm{2}{x}\right)}\right)^{\mathrm{2}} }}.\frac{{cos}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\sqrt{{sin}\left(\mathrm{2}{x}\right)}\right)^{\mathrm{2}} }}.\frac{{cos}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\sqrt{{sin}\left(\mathrm{2}{x}\right)}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sinh}^{−\mathrm{1}} \left(\sqrt{{sin}\left(\mathrm{2}{x}\right)}\right)+{c} \\ $$$$ \\ $$$${i}\:{can}\:{solve}\:{it}\:{with}\:{using}\:{another}\:{ways} \\ $$$${but}\:{i}\:{dont}\:{have}\:{a}\:{time}\:{now}\:{i}\:{will}\:{post} \\ $$$${them}\:{in}\:{anothe}\:{time} \\ $$
Commented by Shakhzod last updated on 12/May/20
Thank you. I will wait you then.
$${Thank}\:{you}.\:{I}\:{will}\:{wait}\:{you}\:{then}. \\ $$
Answered by  M±th+et+s last updated on 12/May/20
Answered by  M±th+et+s last updated on 12/May/20
Answered by  M±th+et+s last updated on 12/May/20
=(1/( (√2)))∫((2sin(x))/( (√(sin(2x)))))dx=(1/( (√2)))∫((cos(x)+sin(x)−cos(x)+sin(x))/( (√(sin(2x)))))dx =(1/( (√2)))∫(((√(1+sin(2x)))−(√(1−sin(2x))))/( (√(sin(2x))))).((csc(2x)cot(2x))/(csc(2x)cot(2x)))dx =(1/( (√2)))∫((((√(csc(2x)+1))−(√(csc(2x)−1))))/( (√(csc^2 (2x)−1))))cot(2x)dx (1/( (√2)))∫((cot(2x))/( (√(csc(2x)−1))))dx−(1/( (√2)))∫((cot(2x))/( (√(csc(2x)+1))))dx (1/( (√2)))∫(((csc(2x)cot(2x))/( (√(csc(2x)−1))))/(((√(csc2x−1)))^2 +1))dx−(1/( (√2)))∫(((csc(2x)cot(2x))/( (√(csc(2x)+1))))/(((√(csc(2x)+1)))^2 −1)) =−(1/( (√2)))tan^(−1) ((√(csc(2x)−1)))−(1/( (√2)))tanh^(−1) ((√(csc(2x)+1)))+c
$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{2}{sin}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{cos}\left({x}\right)+{sin}\left({x}\right)−{cos}\left({x}\right)+{sin}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\sqrt{\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)}−\sqrt{\mathrm{1}−{sin}\left(\mathrm{2}{x}\right)}}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}.\frac{{csc}\left(\mathrm{2}{x}\right){cot}\left(\mathrm{2}{x}\right)}{{csc}\left(\mathrm{2}{x}\right){cot}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\left(\sqrt{{csc}\left(\mathrm{2}{x}\right)+\mathrm{1}}−\sqrt{{csc}\left(\mathrm{2}{x}\right)−\mathrm{1}}\right)}{\:\sqrt{{csc}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{1}}}{cot}\left(\mathrm{2}{x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{cot}\left(\mathrm{2}{x}\right)}{\:\sqrt{{csc}\left(\mathrm{2}{x}\right)−\mathrm{1}}}{dx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{cot}\left(\mathrm{2}{x}\right)}{\:\sqrt{{csc}\left(\mathrm{2}{x}\right)+\mathrm{1}}}{dx} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\frac{{csc}\left(\mathrm{2}{x}\right){cot}\left(\mathrm{2}{x}\right)}{\:\sqrt{{csc}\left(\mathrm{2}{x}\right)−\mathrm{1}}}}{\left(\sqrt{{csc}\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}}{dx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\frac{{csc}\left(\mathrm{2}{x}\right){cot}\left(\mathrm{2}{x}\right)}{\:\sqrt{{csc}\left(\mathrm{2}{x}\right)+\mathrm{1}}}}{\left(\sqrt{{csc}\left(\mathrm{2}{x}\right)+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\sqrt{{csc}\left(\mathrm{2}{x}\right)−\mathrm{1}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tanh}^{−\mathrm{1}} \left(\sqrt{{csc}\left(\mathrm{2}{x}\right)+\mathrm{1}}\right)+{c} \\ $$

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