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Question Number 69734 by mhmd last updated on 27/Sep/19

$$\mathrm{If}{I}_n=\underset{0}{\stackrel{\pi /4}{\int }}{\tan }^{n}xdx,n\in N,\mathrm{then}{I}_{n+2}+I_n=$$

Commented by mathmax by abdo last updated on 16/Oct/19

$$I_{n+2}+I_n={\int }_0^{\frac{\pi }{4}}({tan}^{n+2}x+{tan}^{n}x)dx={\int }_0^{\frac{\pi }{4}}{tan}^{n}x(1+{tan}^{2}x)dx$$$$={\left[\frac{1}{n+1}{tan}^{n+1}x\right]}_0^{\frac{\pi }{4}}=\frac{1}{n+1}$$

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