Menu Close

with-out-special-function-find-0-pi-10-tanx-dx-

Tinku Tara 0 Comments
Pin




Question Number 140628 by mohammad17 last updated on 10/May/21
with out special function find ∫_0 ^( (π/(10))) (√(tanx))dx ?
$$\:{with}\:{out}\:{special}\:{function}\:{find} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{10}}} \sqrt{{tanx}}{dx}\:? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 10/May/21
∫(√(tanx)) dx tanx=t^2 =∫((2t^2 )/(1+t^4 ))dt=2∫(1/((t^2 +(1/t^2 ))))dt=∫((1−(1/t^2 ))/((t+(1/t))^2 −2))+((1+(1/t^2 ))/((t−(1/t))^2 +2))dt =(1/( 2(√2)))log(((t+(1/t)−(√2))/(t+(1/t)+(√2))))+(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2))))+C =(1/(2(√2)))log(((t^2 −(√2)t+1)/(t^2 +(√2)t+1)))+(1/( (√2)))tan^(−1) (((t^2 −1)/( (√(2t)))))+C ∫_0 ^(π/(10)) (√(tanx)) dx=(1/(2(√2)))log(((tan(π/(10))−(√(2tan(π/(10))))+1)/(tan(π/(10))+(√(2tan(π/(10))))+1)))+(1/( (√2)))tan^(−1) (((tan(π/(10))−1)/( (√(2tan(π/(10)))))))+(π/(2(√2))) tan(π/(10))=(((√5)−1)/( (√(10+2(√5))))) =(1/( (√2)))log((((√5)−1)/(10+2(√5)))−(√(((√5)−1)/(5+(√5))))+1)−(1/(2(√2)))log((8/(5+(√5))))+(1/( (√2)))tan^(−1) ((((((√5)−1)/( (√(10+2(√5)))))−1)/( (√(((√5)−1)/(5+(√5)))))))+(π/(2(√2)))
$$\int\sqrt{{tanx}}\:{dx}\:\:\:\:{tanx}={t}^{\mathrm{2}} \\ $$$$=\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}=\mathrm{2}\int\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{dt}=\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{2}{t}}}\right)+{C} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{10}}} \sqrt{{tanx}}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{{tan}\frac{\pi}{\mathrm{10}}−\sqrt{\mathrm{2}{tan}\frac{\pi}{\mathrm{10}}}+\mathrm{1}}{{tan}\frac{\pi}{\mathrm{10}}+\sqrt{\mathrm{2}{tan}\frac{\pi}{\mathrm{10}}}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{\pi}{\mathrm{10}}−\mathrm{1}}{\:\sqrt{\mathrm{2}{tan}\frac{\pi}{\mathrm{10}}}}\right)+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${tan}\frac{\pi}{\mathrm{10}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{log}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{5}+\sqrt{\mathrm{5}}}}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{\mathrm{8}}{\mathrm{5}+\sqrt{\mathrm{5}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}−\mathrm{1}}{\:\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{5}+\sqrt{\mathrm{5}}}}}\right)+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Answered by Dwaipayan Shikari last updated on 10/May/21
∫_0 ^a sin^α (x)cos^β (x)dx =∫_0 ^(sin(a)) (t^α /( (√(1−t^2 ))))((√(1−t^2 )))^β dt =∫_0 ^(sin(a)) (t^α /((1−t^2 )^((1−β)/2) ))dt=Σ_(n≥0) ∫_0 ^(sin(a)) (((((1−β)/2))_n )/(n!))t^(α+2n) dt =sin^(α+1) (a)Σ_(n≥0) (((((1−β)/2))_n )/(n!(α+2n+1)))(sin^2 (a))^(2n) =(1/(α+1))sin^(α+1) (a)Σ_(n≥0) (((((1−β)/2))_n (((α+1)/2))_n )/(n!(((α+3)/2))_n ))(sin^2 (a))^(2n) =((sin^(α+1) (a))/(α+1)) _2 F_1 (((1−β)/2),((α+1)/2);((α+3)/2);sin^2 (a)) ∫_0 ^(π/(10)) (√(tanx)) dx=((sin^(3/2) ((π/(10))))/((1/2)+1)) _2 F_1 ((3/4),(3/4);(7/4);((3−(√5))/8)) =(2/3)(((3−(√5))/8))^(3/2) _2 F_1 ((3/4);(3/4);(7/4);((3−(√5))/8))
$$\int_{\mathrm{0}} ^{{a}} {sin}^{\alpha} \left({x}\right){cos}^{\beta} \left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{{sin}\left({a}\right)} \frac{{t}^{\alpha} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\left(\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)^{\beta} {dt} \\ $$$$=\int_{\mathrm{0}} ^{{sin}\left({a}\right)} \frac{{t}^{\alpha} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}−\beta}{\mathrm{2}}} }{dt}=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{{sin}\left({a}\right)} \frac{\left(\frac{\mathrm{1}−\beta}{\mathrm{2}}\right)_{{n}} }{{n}!}{t}^{\alpha+\mathrm{2}{n}} {dt} \\ $$$$={sin}^{\alpha+\mathrm{1}} \left({a}\right)\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}−\beta}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\alpha+\mathrm{2}{n}+\mathrm{1}\right)}\left({sin}^{\mathrm{2}} \left({a}\right)\right)^{\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{1}}{\alpha+\mathrm{1}}{sin}^{\alpha+\mathrm{1}} \left({a}\right)\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}−\beta}{\mathrm{2}}\right)_{{n}} \left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\frac{\alpha+\mathrm{3}}{\mathrm{2}}\right)_{{n}} }\left({sin}^{\mathrm{2}} \left({a}\right)\right)^{\mathrm{2}{n}} \\ $$$$=\frac{{sin}^{\alpha+\mathrm{1}} \left({a}\right)}{\alpha+\mathrm{1}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}−\beta}{\mathrm{2}},\frac{\alpha+\mathrm{1}}{\mathrm{2}};\frac{\alpha+\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \left({a}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{10}}} \sqrt{{tanx}}\:{dx}=\frac{{sin}^{\mathrm{3}/\mathrm{2}} \left(\frac{\pi}{\mathrm{10}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{7}}{\mathrm{4}};\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}\right)^{\mathrm{3}/\mathrm{2}} \:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{7}}{\mathrm{4}};\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}\right) \\ $$

Pin

Leave a Reply

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