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Question Number 105781 by IE last updated on 31/Jul/20

$$\mathrm{Please},\mathrm{I}\mathrm{need}\mathrm{help}.$$$$\mathrm{Exercise}$$$$\mathrm{We}\mathrm{have}:$$$${\mathrm{J}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{4}}{\tan }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$$$$$1)\mathrm{Establish}\mathrm{a}\mathrm{recurrence}\mathrm{relation}$$$$\mathrm{between}{\mathbf{J}}_{\mathbf{n}+2}\mathrm{and}{\mathbf{J}}_{\mathbf{n}}.$$$$2)\mathrm{Calculate}{\mathbf{J}}_0\mathrm{and}{\mathbf{J}}_1,\mathrm{then}$$$$\mathrm{deduce}\mathrm{the}\mathrm{expression}\mathrm{of}{\mathbf{J}}_{\mathbf{n}}\mathrm{as}\mathrm{a}$$$$\mathrm{function}\mathrm{of}\mathbf{n}.$$$$\mathrm{The}\mathrm{deduction}\mathrm{of}\mathrm{the}\mathrm{last}$$$$\mathrm{question},\mathrm{please}.$$

Commented by prakash jain last updated on 31/Jul/20

$$q2088fornodd$$

Answered by prakash jain last updated on 31/Jul/20

$${\int }_0^{\pi /4}{\tan }^{n}xdx={\int }_0^{\pi /4}{\tan }^{n-2}x({\sec }^{2}x-1)dx$$$$={\int }_0^{\pi /4}{\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\pi /4}{\tan }^{n-2}xdx$$$$I_n={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\pi /4}-I_{n-2}$$$$I_n=\frac{1}{n-1}-I_{n-2}$$$$I_0={\int }_0^{\pi /4}dx=\frac{\pi }{4}$$$$I_1={\int }_0^{\pi /4}\tan xdx=\frac{\ln 2}{2}$$$$I_2=1-I_0$$$$I_3=\frac{1}{2}-I_1$$$$I_4=\frac{1}{3}-1+I_0$$$$I_6=\frac{1}{5}-\frac{1}{3}+1-I_0$$$$I_{2n}={(-1)}^n[\frac{\pi }{4}+\underset{m=1}{\sum ^n}\frac{{(-1)}^m}{2m-1}]$$$$\mathrm{Question}2088\mathrm{for}n\mathrm{odd}(2n+1)$$

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

$$\mathrm{Even}n\mathrm{formula}\mathrm{is}\mathrm{above}.\mathrm{Odd}$$$$n\mathrm{i}\mathrm{have}\mathrm{mentioned}\mathrm{the}\mathrm{question}$$$$\mathrm{where}\mathrm{it}\mathrm{is}\mathrm{derived}$$$$I_{2n+1}={(-1)}^n(\frac{1}{2}ln2+\underset{m=1}{\sum ^n}\frac{{(-1)}^m}{2m})$$

Commented by IE last updated on 31/Jul/20

$$\mathrm{Thanks}!$$

Commented by IE last updated on 31/Jul/20

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!\mathrm{Show}\mathrm{me}$$$$\mathrm{the}\mathrm{summation}\mathrm{formula},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{curious}$$$$\mathrm{to}\mathrm{see}\mathrm{what}\mathrm{it}\mathrm{might}\mathrm{look}\mathrm{like}.$$

Commented by 1549442205PVT last updated on 01/Aug/20

$$\mathrm{Why}\mathrm{do}\mathrm{Sir}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{use}\mathrm{the}\mathrm{formular}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{\mathrm{n}-1}-{\mathrm{I}}_{\mathrm{n}-2}\mathrm{to}\mathrm{calculate}\mathrm{in}\mathrm{the}\mathrm{case}$$$$\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{like}\mathrm{as}\mathrm{did}{\mathrm{forI}}_5.\mathrm{Please},\mathrm{sir}\mathrm{can}$$$$\mathrm{explain}?$$

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