MCQ
${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} $ then $\mathop {\lim }\limits_{n \to \infty } \,\,n({I_n} + {I_{n - 2}})$ equals
- A$1/2$
- ✓$1$
- C$\infty$
- D$0$
$=\int_{0}^{\pi / 4}\left(\tan ^{n-2} x\left(1+\tan ^{2} x\right) d x\right.$
$=\int_{0}^{\pi / 4} \tan ^{n-2} x \sec ^{2} x d x$
$=\left[\frac{\tan ^{\mathrm{n}-1} \mathrm{x}}{\mathrm{n}-1}\right]_{0}^{\pi / 4}$
$ = {I_n} + {I_{n - 2}} = \frac{1}{{n - 1}}\quad \mathop {\lim }\limits_{x \to \infty } \frac{n}{{n - 1}} = 1$
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