options 3/π log2 2/πlog2 3/π log3 3/π

  1. 3/π log2
  2. 2/πlog2
  3. 3/π log3
  4. 3/π

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Hello Student,
Please find answer to your question below

L = \lim_{n\rightarrow \infty }[tan\frac{\pi }{3n}+tan\frac{2\pi }{3n}+tan\frac{3\pi }{3n}+.................+tan\frac{\pi }{3}]\frac{1}{n}
L = \lim_{n\rightarrow \infty }[\sum_{r = 1}^{n}tan\frac{r\pi }{3n}]\frac{1}{n}
L = \lim_{n\rightarrow \infty }[\sum_{r = 1}^{n}\frac{1}{n}.tan\frac{r\pi }{3n}]
Now we are going to convert it into definite integral using the definition
\frac{r}{n} = x,\sum_{r = 1}^{n}\frac{1}{n} = dx
Limit is from 0 to 1.
L = \int_{0}^{1}tan\frac{\pi x}{3}dx
L = \int_{0}^{1}\frac{sin\frac{\pi x}{3}}{cos\frac{\pi x}{3}}dx
L = \frac{3}{\pi }\int_{0}^{1}\frac{\frac{\pi }{3}.sin\frac{\pi x}{3}}{cos\frac{\pi x}{3}}dx
cos\frac{\pi x}{3} = t
-\frac{\pi }{3}sin\frac{\pi x}{3}dx = dt
x = 0\rightarrow t = 1
x = 1\rightarrow t = \frac{1}{2}
L = \frac{3}{\pi }\int_{1}^{1/2}\frac{-1}{t}dt
L = \frac{3}{\pi }\int_{1/2}^{1}\frac{1}{t}dt
L = \frac{3}{\pi }[logt]_{1/2}^{1}
L = \frac{3}{\pi }[-log(1/2)]
L = \frac{3}{\pi }[log(2)]
Option (1) is correct.

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