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Question Number 116196 by mnjuly1970 last updated on 01/Oct/20

$$pleasesolve:$$$$$$$${\int }_0^{\frac{\pi }{4}}{tan}^9\left(x\right)dx=???$$

Answered by mindispower last updated on 01/Oct/20

$$let$$$$u_n={\int }_0^{\frac{\pi }{4}}{tg}^n\left(x\right)dx$$$$U_{n+2}+U_n={\int }_0^{\frac{\pi }{4}}{tg}^n\left(x\right)dx+{\int }_0^{\frac{\pi }{4}}{tg}^{n+2}\left(x\right)dx$$$$={\int }_0^{\frac{\pi }{4}}{tg}^n\left(x\right)(1+{tg}^2\left(x\right))dx={\int }_0^{\frac{\pi }{4}}{tg}^n\left(x\right)d\left(tg\left(x\right)\right)$$$$={\left[\frac{{tg}^{n+1}\left(x\right)}{n+1}\right]}_0^{\frac{\pi }{4}}=\frac{1}{n+1}$$$$\Rightarrow U_{n+2}=\frac{1}{n+1}-U_n$$$$U_0=\frac{\pi }{4},U_1={\int }_0^{\frac{\pi }{4}}tg\left(x\right)dx={[-ln\left(cos\left(x\right)\right)]}_0^{\frac{\pi }{4}}$$$$=ln\left(\sqrt{2}\right)$$$$U_3=-U_1+\frac{1}{2}$$$$U_5=-U_3+\frac{1}{4}$$$$U_7=-U_5+\frac{1}{6}$$$$U_9=-U_7+\frac{1}{8}$$$$=\frac{1}{8}-\frac{1}{6}+\frac{1}{4}-\frac{1}{2}+ln\left(\sqrt{2}\right)$$$$=\frac{3-4+6-12}{24}+ln\left(\sqrt{2}\right)$$$$=\frac{-7+12ln\left(2\right)}{24}$$

Commented by mnjuly1970 last updated on 01/Oct/20

$$thankyousir..$$

Answered by MJS_new last updated on 01/Oct/20

$$\int {\tan }^{9}xdx=$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=\int \frac{t^9}{t^2+1}dt=\int (\frac{t}{t^2+1}+t^7-t^5+t^3-t)dt=$$$$=\frac{1}{2}\ln (t^2+1)+\frac{t^8}{8}-\frac{t^6}{6}+\frac{t^4}{4}-\frac{t^2}{2}$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{1}{2}\ln 2-\frac{7}{24}$$

Commented by mnjuly1970 last updated on 01/Oct/20

$$grateful..$$

Answered by mathmax by abdo last updated on 01/Oct/20

$$\mathrm{let}{\mathrm{u}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{4}}{\tan }^{2\mathrm{n}+1}\mathrm{xdx}\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{4}}{\tan }^{2\mathrm{n}-1}\mathrm{x}(1+{\tan }^2\mathrm{x}-1)\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{4}}{\tan }^{2\mathrm{n}-1}(1+{\tan }^2\mathrm{x})\mathrm{dx}-{\mathrm{u}}_{\mathrm{n}-1}\mathrm{we}\mathrm{hsve}$$$${\int }_0^{\frac{\pi }{4}}{(1+{\tan }^2\mathrm{x})}^{2\mathrm{n}-1}\mathrm{dx}={\left[\frac{1}{2\mathrm{n}}{\tan }^{2\mathrm{n}}\mathrm{x}\right]}_0^{\frac{\pi }{4}}=\frac{1}{2\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{1}{2\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}-1}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}-1}=\frac{1}{2\mathrm{n}}\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}{(-1)}^{\mathrm{k}}({\mathrm{u}}_{\mathrm{k}}+{\mathrm{u}}_{\mathrm{k}-1})=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\Rightarrow$$$$-({\mathrm{u}}_1+{\mathrm{u}}_0)+({\mathrm{u}}_2+{\mathrm{u}}_1)+...{(-1)}^{\mathrm{n}-1}({\mathrm{u}}_{\mathrm{n}-1}+{\mathrm{u}}_{\mathrm{n}-2})+{(-1)}^{\mathrm{n}}({\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}-1})$$$$=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\Rightarrow -{\mathrm{u}}_0+{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\Rightarrow$$$${(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_0+\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}$$$${\mathrm{u}}_0={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}={[-\ln \mid \mathrm{cosx}\mid ]}_0^{\frac{\pi }{4}}=-\ln \left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}\ln \left(2\right)\Rightarrow$$$${(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{2}\ln \left(2\right)+\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{4}}{\tan }^{2\mathrm{n}+1}\mathrm{x}\mathrm{dx}={(-1)}^{\mathrm{n}}\{\frac{\ln 2}{2}+\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\}$$$$\mathrm{n}=4\Rightarrow {\int }_0^{\frac{\pi }{4}}{\tan }^9\mathrm{xdx}=\frac{\ln 2}{2}+\frac{1}{2}\sum _{\mathrm{k}=1}^4\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}$$$$=\frac{\ln 2}{2}+\frac{1}{2}\{-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\}=\frac{\ln 2}{2}+\frac{1}{2}\{-\frac{4}{3}+\frac{3}{4}\}$$$$=\frac{\ln 2}{2}+\frac{1}{2}\left(\frac{-7}{12}\right)=\frac{\ln 2}{2}-\frac{7}{24}$$

Answered by Dwaipayan Shikari last updated on 01/Oct/20

$${\int }_0^{\frac{\pi }{4}}\frac{{\sin }^9\mathrm{x}}{{\cos }^9\mathrm{x}}\mathrm{dx}=-{\int }_0^{\frac{\pi }{4}}\frac{{\sin }^8\mathrm{x}(-\mathrm{sinx})}{{\cos }^9\mathrm{x}}\mathrm{dx}$$$$-{\int }_1^{\frac{1}{\sqrt{2}}}\frac{{(1-{\mathrm{t}}^2)}^4}{{\mathrm{t}}^9}\mathrm{dt}={\int }_{\frac{1}{\sqrt{2}}}^1\frac{1}{{\mathrm{t}}^9}-\frac{4}{{\mathrm{t}}^7}+\frac{6}{{\mathrm{t}}^5}-\frac{4}{{\mathrm{t}}^3}+\frac{1}{\mathrm{t}}$$$$={[-\frac{1}{{8\mathrm{t}}^8}+\frac{2}{{3\mathrm{t}}^6}-\frac{3}{{2\mathrm{t}}^4}+\frac{2}{{\mathrm{t}}^2}+\log \left(\mathrm{t}\right)]}_{\frac{1}{\sqrt{2}}}^1=\log \left(\sqrt{2}\right)-\frac{7}{24}$$$$$$$$$$

Answered by maths mind last updated on 01/Oct/20

$${\int }_0^{\frac{\pi }{4}}{tg}^a\left(x\right)dx=f\left(a\right),a\in \mathbb{C}Re\left(a\right)⩾0$$$$tg\left(x\right)=t\Rightarrow f\left(a\right)={\int }_0^1\frac{t^a}{1+t^2}dt={\int }_0^1\sum _{m⩾0}{(-t^2)}^m{t}^{a}dt$$$$=\sum _{m⩾0}{(-1)}^m{\int }_0^1{t}^{2m+a}dt$$$$=\sum _{m⩾0}\frac{{(-1)}^m}{2m+a+1}=\sum _{m⩾0}(\frac{1}{4m+a+1}-\frac{1}{4m+a+3})$$$$=\frac{1}{4}\sum _{m⩾0}\left(\frac{\frac{a+3}{4}-\frac{a+1}{4}}{(m+\frac{a+1}{4})(m+\frac{a+2}{4})}\right)=\frac{2}{16}.\frac{\mathrm{\Psi }\left(\frac{a+3}{4}\right)-\mathrm{\Psi }\left(\frac{a+1}{4}\right)}{\frac{a+3}{4}-\frac{a+1}{4}}$$$$=\frac{\mathrm{\Psi }\left(\frac{a+3}{4}\right)-\mathbf{\Psi }\left(\frac{a+1}{4}\right)}{4}=f\left(a\right)$$$$f\left(9\right)=\frac{\mathrm{\Psi }\left(\frac{12}{4}\right)-\mathrm{\Psi }\left(\frac{10}{4}\right)}{4}$$$$\mathrm{\Psi }(z+1)=\mathrm{\Psi }\left(z\right)+\frac{1}{z}$$$$\mathrm{\Psi }\left(3\right)=-\gamma +1+\frac{1}{2}=\frac{3}{2}-\gamma$$$$\mathrm{\Psi }\left(\frac{10}{4}\right)=\mathrm{\Psi }\left(\frac{5}{2}\right)=\mathrm{\Psi }\left(\frac{3}{2}\right)+\frac{2}{3}=\frac{2}{3}+2+\mathrm{\Psi }\left(\frac{1}{2}\right),\mathrm{\Psi }\left(\frac{1}{2}\right)=-2ln\left(2\right)-\gamma$$$$\mathrm{\Psi }\left(3\right)-\mathrm{\Psi }\left(\frac{5}{2}\right)=\frac{3}{2}-\gamma -\frac{2}{3}-2-\mathrm{\Psi }\left(\frac{1}{2}\right)=\frac{3}{2}-\gamma -\frac{8}{3}+2ln\left(2\right)+\gamma$$$$=\frac{-7}{6}+2ln\left(2\right)$$$$f\left(9\right)=\frac{-7+12ln\left(2\right)}{24}=-\frac{7}{24}+\frac{ln\left(2\right)}{2}$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 02/Oct/20

$$verynice..thankyouso$$$$much....$$

Answered by 1549442205PVT last updated on 02/Oct/20

$${\mathrm{I}}_9=\int {\tan }^9\mathrm{xdx}=\int {\tan }^7\mathrm{x}(1+{\tan }^2\mathrm{x})\mathrm{dx}$$$$-\int {\tan }^7\mathrm{xdx}=\int {\tan }^7\mathrm{xd}\left(\mathrm{tanx}\right)-{\mathrm{I}}_7$$$$=\frac{{\tan }^8\mathrm{x}}{8}-{\mathrm{I}}_7=\frac{{\tan }^8\mathrm{x}}{8}-\frac{{\tan }^6\mathrm{x}}{6}+\frac{{\tan }^4\mathrm{x}}{4}$$$$-\frac{{\tan }^2\mathrm{x}}{2}+{\mathrm{I}}_1=\frac{{\tan }^8\mathrm{x}}{8}-\frac{{\tan }^6\mathrm{x}}{6}+\frac{{\tan }^4\mathrm{x}}{4}-\frac{{\tan }^2\mathrm{x}}{2}-\mathrm{lncosx}$$$$\mathrm{Hence},{\int }_0^{\frac{\pi }{4}}{tan}^9\left(x\right)dx={\mathrm{I}}_9{\mid }_0^{\pi /4}$$$$=\frac{1}{8}-\frac{1}{6}+\frac{1}{4}-\frac{1}{2}-\ln \frac{1}{\sqrt{2}}=\frac{-7}{24}+\frac{1}{2}\ln 2$$

Commented by mnjuly1970 last updated on 02/Oct/20

$$thanksalot$$

Commented by 1549442205PVT last updated on 07/Oct/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}.$$

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