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Question Number 78458 by kaivan.ahmadi last updated on 17/Jan/20

$$if{I}_n={\int }_0^{\frac{\pi }{4}}{tan}^{n}xdxand{I}_n=a_n+b_n{I}_{n-2}thenfind{a}_{10}$$

Commented by kaivan.ahmadi last updated on 17/Jan/20

$$I_n+I_{n-2}=\int {tg}^{n}xdx+\int {tg}^{n-2}xdx=\int ({tg}^{n}xdx+{tg}^{n-2}x)dx$$$$=\int {tg}^{n-2}x({tg}^{2}x+1)dx$$$$u=tgx\Rightarrow du=(1+{tg}^{2}x)dx\Rightarrow \int u^{n-2}du=$$$$\frac{u^{n-3}}{n-1}=\frac{{tg}^{n-3}x}{n-1}{\mid }_0^{\frac{\pi }{4}}=\frac{1}{n-1}$$$$\Rightarrow$$$$$$$$I_n+I_{n-2}=\frac{1}{n-1}$$$$\Rightarrow I_n=\frac{1}{n-1}-I_{n-2}=a_n+(-1)I_{n-2}\Rightarrow a_n=\frac{1}{n-1}$$$$\Rightarrow a_{10}=\frac{1}{9}$$$$$$

Commented by mathmax by abdo last updated on 17/Jan/20

$$I_n={\int }_0^{\frac{\pi }{4}}{tan}^{n}xdx\Rightarrow I_n={\int }_0^{\frac{\pi }{4}}{tan}^{n-2}x(1+{tan}^{2}x-1)dx$$$$={\int }_0^{\frac{\pi }{4}}(1+{tan}^{2}x){tan}^{n-2}xdx-{\int }_0^{\frac{\pi }{4}}{tan}^{n-2}xdx$$$$={\left[\frac{1}{n-1}{tan}^{n-1}x\right]}_0^{\frac{\pi }{4}}-I_{n-2}=\frac{1}{n-1}-I_{n-2}ifn⩾2\Rightarrow a_n=\frac{1}{n-1}$$$$and{b}_n=-1\Rightarrow a_{10}=\frac{1}{9}$$$$letcalculate{I}_{n}intermsofnwehave$$$$I_n+I_{n-2}=\frac{1}{n-1}\Rightarrow I_{2n}+I_{2n-2}=\frac{1}{2n-1}let{I}_{2n}=u_n\Rightarrow$$$$u_n+u_{n-1}=\frac{1}{2n-1}\Rightarrow \sum _{k=1}^n{(-1)}^k(u_k+u_{k-1})=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow$$$$-(u_1+u_0)+u_2+u_1+....{(-1)}^n(u_n+u_{n-1})=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}$$$${(-1)}^n{u}_n-u_0=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow {(-1)}^n{u}_n=u_0+\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow$$$$u_n={(-1)}^n{u}_0+{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}=\frac{\pi }{4}{(-1)}^n+{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}$$$$\Rightarrow I_{2n}=\frac{\pi }{4}{(-1)}^n+{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}letdetemine{I}_{2n+1}?$$$$wehave{I}_{2n+1}+I_{2n-1}=\frac{1}{2n}\Rightarrow \sum _{k=1}^n{(-1)}^k(v_k+v_{k-1})=\sum _{k=1}^n\frac{{(-1)}^k}{2k}$$$$(v_n=I_{2n+1})\Rightarrow {(-1)}^n{v}_n-v_0=\sum _{k=1}^n\frac{{(-1)}^k}{2k}\Rightarrow$$$${(-1)}^n{v}_n=v_0+\sum _{k=1}^n\frac{{(-1)}^k}{2k}\Rightarrow$$$$v_n={(-1)}^n{v}_0+{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k}$$$${v_0=I_1={\int }_0^{\frac{\pi }{4}}tanxdx=-{\int }_0^{\frac{\pi }{4}}\frac{-sinx}{cosx}dx=-ln\mid cosx\mid ]}_0^{\frac{\pi }{4}}$$$$=-ln\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}ln\left(2\right)\Rightarrow I_{2n+1}={(-1)}^n\frac{ln\left(2\right)}{2}+{(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k}$$

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