A calculus problem by A Former Brilliant Member

Calculus Level 4

Let I n = 0 π / 4 tan n x d x \displaystyle I_n = \int_0^{\pi /4} \tan^n x \, dx . Find lim n n ( I n + I n + 2 ) \displaystyle \lim_{n\to\infty} n ( I_n + I_{n+2} ) .

\infty It depends on n n 1 2 \frac12 1 -1 1 1 2 2 0 0

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1 solution

lim n n ( I n + I n + 2 ) = lim n n ( 0 π 4 tan n x d x + 0 π 4 tan n + 2 x d x ) = lim n n 0 π 4 tan n x ( 1 + tan 2 x ) d x = lim n n 0 π 4 tan n x sec 2 x d x Let t = tan x d t = sec 2 x d x = lim n n 0 1 t n d t = lim n n t n + 1 n + 1 0 1 = lim n n n + 1 = lim n 1 1 + 1 n = 1 \begin{aligned} \lim_{n \to \infty} n(I_n + I_{n+2}) & = \lim_{n \to \infty} n \left(\int_0^\frac \pi 4 \tan^n x \ dx + \int_0^\frac \pi 4 \tan^{n+2} x \ dx \right) \\ & = \lim_{n \to \infty} n \int_0^\frac \pi 4 \tan^n x (1+\tan^2 x) \ dx \\ & = \lim_{n \to \infty} n \int_0^\frac \pi 4 \tan^n x \sec^2 x \ dx & \small \color{#3D99F6}{\text{Let }t = \tan x \implies dt = \sec^2 x \ dx} \\ & = \lim_{n \to \infty} n \int_0^1 t^n \ dt \\ & = \lim_{n \to \infty} n \cdot \frac {t^{n+1}}{n+1} \bigg|_0^1 \\ & = \lim_{n \to \infty} \frac n{n+1} \\ & = \lim_{n \to \infty} \frac 1{1+\frac 1n} \\ & = \boxed{1} \end{aligned}

Perfect solution, I did the same way as well, Nice question Shubham.

Swagat Panda - 4 years, 10 months ago

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That is why it is perfect.

Chew-Seong Cheong - 4 years, 10 months ago

most pleased to hear that brother.

A Former Brilliant Member - 4 years, 10 months ago

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This question was asked in one of our definite integral RRQ's , if I am not mistaken.

Swagat Panda - 4 years, 10 months ago

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@Swagat Panda no it's i module i don't know about rrq much ? how is your advance 3 preparation going ?

A Former Brilliant Member - 4 years, 10 months ago

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