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

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(A) दिया गया है ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. हम लिख सकते हैं ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{	an ^n}x\,dx} + \int_0^{\pi /4}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) दिया गया है ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. हम लिख सकते हैं ${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} $. चूंकि $	an ^2x + 1 = \sec ^2x$,इसलिए: $= \int_0^{\pi /4} {{	an ^{n - 2}}x\sec ^2x\,dx} $. मान लीजिए $t = 	an x$,तो $dt = \sec ^2x\,dx$. जब $x = 0, t = 0$ और जब $x = \pi /4, t = 1$. $= \int_0^1 {{t^{n - 2}}\,dt} $. $= \left[ \frac{t^{n - 1}}{n - 1} ight]_0^1 = \frac{1}{n - 1} - 0 = \frac{1}{n - 1}$. अतः,${u_n} + {u_{n - 2}} = \frac{1}{n - 1}$.

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

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