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Question Number 123552 by Lordose last updated on 26/Nov/20

$${\int }_0^{\frac{\pi }{4}}{\tan }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by Adeleke last updated on 26/Nov/20

Answered by TANMAY PANACEA last updated on 26/Nov/20

$$I_n={\int }_0^{\frac{\pi }{4}}{tan}^{n-2}x({sec}^{2}x-1)dx$$$$={\int }_0^{\frac{\pi }{4}}{tan}^{n-2}x\times d\left(tanx\right)-I_{n-2}$$$$=\mid \frac{{tan}^{n-1}x}{n-1}{\mid }_0^{\frac{\pi }{4}}-I_{n-2}$$$$=\frac{1}{n-1}-I_{n-2}$$$$I_n=\frac{1}{n-1}-I_{n-2}$$

Commented by peter frank last updated on 26/Nov/20

$$\mathrm{thank}\mathrm{you}$$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

$$Ifitwas{\int }_0^{\frac{\pi }{2}}{tan}^{n}xdx$$$$thenwecangetaclosedform$$$${\int }_0^{\frac{\pi }{2}}{sin}^{n}x{cos}^{-n}xdx$$$$=\frac{1}{2}{\int }_0^1{t}^{\frac{n}{2}-\frac{1}{2}}{(1-t)}^{\frac{-n}{2}-\frac{1}{2}}dt{sin}^{2}x=t$$$$=\frac{1}{2}\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})\mathrm{\Gamma }(\frac{1}{2}-\frac{n}{2})}{\mathrm{\Gamma }\left(1\right)}=\frac{\pi }{2sin(\frac{n\pi }{2}+\frac{\pi }{2})}$$$${\int }_0^{\frac{\pi }{4}}{tan}^{p}xdx={\int }_0^1\frac{t^p}{t^2+1}dt$$$$={\int }_0^1{t}^p\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{t}^{2n}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1{t}^{2n+p}dt=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{2n+p+1}$$

Answered by mnjuly1970 last updated on 26/Nov/20

$$\mathrm{\Omega }\stackrel{tan\left(x\right)=y}{=}{\int }_0^1\frac{y^{n}dy}{1+y^2}={\int }_0^1\frac{y^n-y^{n+2}}{1-y^4}dy$$$$=\frac{1}{4}{\int }_0^1\frac{t^{\frac{n-3}{4}}-t^{\frac{n-1}{4}}}{1-t}dt$$$$=\frac{1}{4}({\mathrm{H}}_{\frac{n-1}{4}}-{\mathrm{H}}_{\frac{n-3}{4}})$$$$=\frac{1}{4}(\psi (\frac{n}{4}+\frac{3}{4})-\psi (\frac{n}{4}+\frac{1}{4}))✓$$$$$$$$$$$$$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 26/Nov/20

$${\int }_0^{\frac{\pi }{4}}{tan}^{4}xdx=\frac{1}{4}(\psi (1+\frac{3}{4})-\psi (1+\frac{1}{4}))$$$$=\frac{1}{4}(\psi \left(\frac{3}{4}\right)+\frac{4}{3}-\psi \left(\frac{1}{4}\right)-4)$$$$=\frac{1}{4}(\pi cot\left(\frac{\pi }{4}\right)-\frac{8}{3})=\frac{\pi }{4}-\frac{2}{3}$$

Commented by Lordose last updated on 02/Dec/20

$$\mathrm{Exactly}$$

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