Question Number 105781 by IE last updated on 31/Jul/20
$$\mathrm{Please},\:\mathrm{I}\:\mathrm{need}\:\mathrm{help}. \\ $$$$\mathrm{Exercise} \\ $$$$\mathrm{We}\:\mathrm{have}\:: \\ $$$$\mathrm{J}_{\mathrm{n}} =\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{n}} \left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Establish}\:\mathrm{a}\:\mathrm{recurrence}\:\mathrm{relation} \\ $$$$\mathrm{between}\:\boldsymbol{\mathrm{J}}_{\boldsymbol{\mathrm{n}}+\mathrm{2}} \:\mathrm{and}\:\boldsymbol{\mathrm{J}}_{\boldsymbol{\mathrm{n}}} . \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Calculate}\:\boldsymbol{\mathrm{J}}_{\mathrm{0}} \:\mathrm{and}\:\boldsymbol{\mathrm{J}}_{\mathrm{1}} ,\:\mathrm{then} \\ $$$$\mathrm{deduce}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{of}\:\boldsymbol{\mathrm{J}}_{\boldsymbol{\mathrm{n}}} \:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:\boldsymbol{\mathrm{n}}. \\ $$$$\mathrm{The}\:\mathrm{deduction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last} \\ $$$$\mathrm{question},\:\mathrm{please}. \\ $$
Commented by prakash jain last updated on 31/Jul/20
$${q}\mathrm{2088}\:{for}\:{n}\:{odd} \\ $$
Answered by prakash jain last updated on 31/Jul/20
![∫_0 ^(π/4) tan^n xdx=∫_0 ^(π/4) tan^(n−2) x(sec^2 x−1)dx =∫_0 ^(π/4) tan^(n−2) xsec^2 xdx−∫_0 ^(π/4) tan^(n−2) xdx I_n =[((tan^(n−1) x)/(n−1))]_0 ^(π/4) −I_(n−2) I_n =(1/(n−1))−I_(n−2) I_0 =∫_0 ^(π/4) dx=(π/4) I_1 =∫_0 ^(π/4) tan xdx=((ln 2)/2) I_2 =1−I_0 I_3 =(1/2)−I_1 I_4 =(1/3)−1+I_0 I_6 =(1/5)−(1/3)+1−I_0 I_(2n) =(−1)^n [(π/4)+Σ_(m=1) ^n (((−1)^m )/(2m−1))] Question 2088 for n odd (2n+1)](https://www.tinkutara.com/question/Q105788.png)
$$\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}^{{n}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}^{{n}−\mathrm{2}} {x}\left(\mathrm{sec}^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}^{{n}−\mathrm{2}} {x}\mathrm{sec}^{\mathrm{2}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}^{{n}−\mathrm{2}} {xdx} \\ $$$${I}_{{n}} =\left[\frac{\mathrm{tan}^{{n}−\mathrm{1}} {x}}{{n}−\mathrm{1}}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} −{I}_{{n}−\mathrm{2}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−{I}_{{n}−\mathrm{2}} \\ $$$${I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {dx}=\frac{\pi}{\mathrm{4}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}\:{xdx}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\mathrm{1}−{I}_{\mathrm{0}} \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}−{I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}+{I}_{\mathrm{0}} \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}−{I}_{\mathrm{0}} \\ $$$${I}_{\mathrm{2}{n}} =\left(−\mathrm{1}\right)^{{n}} \left[\frac{\pi}{\mathrm{4}}+\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{m}} }{\mathrm{2}{m}−\mathrm{1}}\right] \\ $$$$\mathrm{Question}\:\mathrm{2088}\:\mathrm{for}\:{n}\:\mathrm{odd}\:\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$
Commented by IE last updated on 31/Jul/20
$$\mathrm{thanks}\:\mathrm{sir}!\:\mathrm{I}\:\mathrm{am}\:\mathrm{waiting}\:\mathrm{more} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{last} \\ $$$$\mathrm{question}\:\left(\mathrm{deduction}\right). \\ $$
Commented by prakash jain last updated on 31/Jul/20
$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{able}\:\mathrm{derive}\:\mathrm{closed} \\ $$$$\mathrm{form}\:\mathrm{formula}.\:\mathrm{Summation} \\ $$$$\mathrm{notation}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{what}\:\mathrm{i}\:\mathrm{have}. \\ $$$$\mathrm{i}\:\mathrm{will}\:\mathrm{continue}\:\mathrm{to}\:\mathrm{think}. \\ $$
Commented by prakash jain last updated on 31/Jul/20
$$\mathrm{Even}\:{n}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{above}.\:\mathrm{Odd} \\ $$$${n}\:\mathrm{i}\:\mathrm{have}\:\mathrm{mentioned}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{where}\:\mathrm{it}\:\mathrm{is}\:\mathrm{derived} \\ $$$${I}_{\mathrm{2}{n}+\mathrm{1}} =\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\mathrm{2}{m}}\right) \\ $$
Commented by IE last updated on 31/Jul/20
$$\mathrm{Thanks}!\:\: \\ $$
Commented by IE last updated on 31/Jul/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}!\:\mathrm{Show}\:\mathrm{me} \\ $$$$\mathrm{the}\:\mathrm{summation}\:\mathrm{formula},\:\mathrm{I}'\mathrm{m}\:\mathrm{curious} \\ $$$$\mathrm{to}\:\mathrm{see}\:\mathrm{what}\:\mathrm{it}\:\mathrm{might}\:\mathrm{look}\:\mathrm{like}. \\ $$
Commented by 1549442205PVT last updated on 01/Aug/20
$$\mathrm{Why}\:\mathrm{do}\:\mathrm{Sir}\:\mathrm{don}'\mathrm{t}\:\mathrm{use}\:\mathrm{the}\:\:\mathrm{formular}\: \\ $$$$\mathrm{I}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}−\mathrm{I}_{\mathrm{n}−\mathrm{2}\:} \mathrm{to}\:\mathrm{calculate}\:\mathrm{in}\:\mathrm{the}\:\mathrm{case} \\ $$$$\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{like}\:\mathrm{as}\:\mathrm{did}\:\mathrm{forI}_{\mathrm{5}} \:.\mathrm{Please},\mathrm{sir}\:\mathrm{can} \\ $$$$\mathrm{explain}\:? \\ $$