7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$ = \int_0^{\pi /4} {{{\sec }^2}x{{\tan }^{n - 2}}} x\,dx - \int_0^{\pi /4} {{{\tan }^{n - 2}}} x\,dx$

$ = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\pi /4} - {I_{n - 2}}$

==> ${I_n} + {I_{n - 2}} = \frac{1}{{n - \,1}}$

Now, $\mathop {\lim }\limits_{n \to \infty } \,n[{I_n} + {I_{n - 2}}]$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{n}{{n - 1}}$

$= \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{1 - \frac{1}{n}}} = 1$.