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Question Number 86454 by redmiiuser last updated on 28/Mar/20
pls check the question below
$$\:{pls}\:{check}\:{the}\: \\ $$$${question}\:{below} \\ $$
Answered by redmiiuser last updated on 28/Mar/20
∫1/(√)(1+x^3 ).dx (1+x^3 )^((−1/2)) =1+(−1/2)x^3 +(−1/2)(−3/2)x^6 /2!+(−1/2)(−3/2)(−5/2)x^9 /3!+(−1/2)(−3/2)(−5/2)(−7/2)/4! x^(12) +... therefore (1+x^3 )^((−1/2)) =Σ_(n=0) ^∞ (((−1)^n .(2n)!.(x^3 )^n )/((2^(2n) .(n!)^2 ))) ∴∫(1+x^3 )^((−1/2)) .dx =∫Σ_(n=0) ^∞ (((−1)^n .(2n)!.(x^3 )^n )/(2^(2n) .(n!)^2 )) .dx =Σ_(n=0) ^∞ (((−1)^n .(2n)!.(x)^(3n+1) )/((3n+1).(2^(2n) ).(n!)^2 ))
$$\int\mathrm{1}/\sqrt{}\left(\mathrm{1}+{x}^{\mathrm{3}} \right).{dx} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\left(−\mathrm{1}/\mathrm{2}\right)} \\ $$$$=\mathrm{1}+\left(−\mathrm{1}/\mathrm{2}\right){x}^{\mathrm{3}} +\left(−\mathrm{1}/\mathrm{2}\right)\left(−\mathrm{3}/\mathrm{2}\right){x}^{\mathrm{6}} /\mathrm{2}!+\left(−\mathrm{1}/\mathrm{2}\right)\left(−\mathrm{3}/\mathrm{2}\right)\left(−\mathrm{5}/\mathrm{2}\right){x}^{\mathrm{9}} /\mathrm{3}!+\left(−\mathrm{1}/\mathrm{2}\right)\left(−\mathrm{3}/\mathrm{2}\right)\left(−\mathrm{5}/\mathrm{2}\right)\left(−\mathrm{7}/\mathrm{2}\right)/\mathrm{4}!\:{x}^{\mathrm{12}} +… \\ $$$$ \\ $$$${therefore} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\left(−\mathrm{1}/\mathrm{2}\right)} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\left(\left(−\mathrm{1}\right)^{{n}} .\left(\mathrm{2}{n}\right)!.\left({x}^{\mathrm{3}} \right)^{{n}} \right)/\left(\left(\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} \right)\right)\right) \\ $$$$\therefore\int\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\left(−\mathrm{1}/\mathrm{2}\right)} .{dx} \\ $$$$=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\left(\left(−\mathrm{1}\right)^{{n}} .\left(\mathrm{2}{n}\right)!.\left({x}^{\mathrm{3}} \right)^{{n}} \right)/\left(\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} \right)\right)\:.{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\left(\left(−\mathrm{1}\right)^{{n}} .\left(\mathrm{2}{n}\right)!.\left({x}\right)^{\mathrm{3}{n}+\mathrm{1}} \right)/\left(\left(\mathrm{3}{n}+\mathrm{1}\right).\left(\mathrm{2}^{\mathrm{2}{n}} \right).\left({n}!\right)^{\mathrm{2}} \right)\right) \\ $$$$ \\ $$
Commented by redmiiuser last updated on 29/Mar/20
can anyone comment whether the above process is correct or wrong.
$${can}\:{anyone}\:{comment} \\ $$$${whether}\:{the}\:{above} \\ $$$${process}\:{is}\:{correct}\:{or} \\ $$$${wrong}. \\ $$
Commented by redmiiuser last updated on 29/Mar/20
mr.Tanmay can you help me in the above problem pls.
$${mr}.{Tanmay}\:{can}\:{you} \\ $$$${help}\:{me}\:{in}\:{the} \\ $$$${above}\:{problem}\:{pls}. \\ $$
Commented by TANMAY PANACEA. last updated on 29/Mar/20
∫(dx/( (√(1+x^3 )) )) x^3 =tan^2 α x=(tanα)^(2/3) →dx=(2/3)×(tanα)^((2/3)−1) ×sec^2 α dα ∫((2(tanα)^((−1)/3) ×sec^2 α×dα)/(3×(1+tan^2 α)^(1/2) )) (2/3)∫((secα dα)/((tanα)^(1/3) )) (2/3)∫(dα/((sinα)^(1/3) (cosα)^(2/3) ))
$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\:} \\ $$$${x}^{\mathrm{3}} ={tan}^{\mathrm{2}} \alpha \\ $$$${x}=\left({tan}\alpha\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}×\left({tan}\alpha\right)^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} ×{sec}^{\mathrm{2}} \alpha\:{d}\alpha \\ $$$$\int\frac{\mathrm{2}\left({tan}\alpha\right)^{\frac{−\mathrm{1}}{\mathrm{3}}} ×{sec}^{\mathrm{2}} \alpha×{d}\alpha}{\mathrm{3}×\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{sec}\alpha\:{d}\alpha}{\left({tan}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{d}\alpha}{\left({sin}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left({cos}\alpha\right)^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$

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