$I_{n} = \int_{0}^{\frac{\pi}{4}}{ an^{n}x}\ dx$, then $\lim_{n ightarrow \infty}{n\lbrack I_{n} + I_{n + 2} brack}$ is equal to

Question

$I_{n} = \int_{0}^{\frac{\pi}{4}}{	an^{n}x}\ dx$, then $\lim_{n ightarrow \infty}{n\lbrack I_{n} + I_{n + 2}brack}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$I_{n} + I_{n + 2} = \int_{0}^{\pi/4}{	an^{n}{x(1 + 	an^{2}{x)dx = \int_{0}^{\pi/4}{	an^{n}{x\sec^{2}{x\ dx}}}}}}$
Put $	an x = t \Rightarrow \sec^{2}{xdx = dt}$
$	herefore I_{n} + I_{n + 2} = \int_{0}^{1}{t^{n}\ dt = \frac{1}{n + 1}}$
$\Rightarrow \lim_{n ightarrow \infty}{n\left\lbrack I_{n} + I_{n + 2} ightbrack = \lim_{n ightarrow \infty}{\frac{n}{n + 1} = \lim_{n ightarrow \infty}{\frac{1}{\left( 1 + \frac{1}{n} ight)^{\ }} = 1}}}$

Q. $I_{n} = \int_{0}^{\frac{\pi}{4}}{\tan^{n}x}\ dx$, then $\lim_{n \rightarrow \infty}{n\lbrack I_{n} + I_{n + 2}\rbrack}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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