If {u_n} = int_0^{pi /4} {{tan ^n}x,dx,} the | vedclass

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(A) Given ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. We can write ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{	an ^n}x\,dx} + \int_0^{\pi /4} {{	an ^

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$\frac{1}{{n - 1}}$

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(A) Given ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. We can write ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{	an ^n}x\,dx} + \int_0^{\pi /4} {{	an ^{n - 2}}x\,dx} $. $= \int_0^{\pi /4} {{	an ^{n - 2}}x(	an ^2x + 1)\,dx} $. Since $	an ^2x + 1 = \sec ^2x$,we have: $= \int_0^{\pi /4} {{	an ^{n - 2}}x\sec ^2x\,dx} $. Let $t = 	an x$,then $dt = \sec ^2x\,dx$. When $x = 0, t = 0$ and when $x = \pi /4, t = 1$. $= \int_0^1 {{t^{n - 2}}\,dt} $. $= \left[ \frac{t^{n - 1}}{n - 1} ight]_0^1 = \frac{1}{n - 1} - 0 = \frac{1}{n - 1}$. Thus,${u_n} + {u_{n - 2}} = \frac{1}{n - 1}$.

If ${u_n} = \int_0^{\pi /4} {{\tan ^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

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