Topics
Examine the convergence of \(\Large\int\limits_{0}^{\pi}\frac{dx}{1+cosx}\).
|
A) convergent |
|
B) converges to 1 |
|
C) converges to 0 |
|
D) diverges |
Correct Answer:
| D) diverges |
Description for Correct answer:
\(\Large \int\limits_{0}^{x}\frac{dx}{1+cosx}\)
when \(x= \pi \) then \(\Large \frac{1}{1+cos \pi }=\infty\)
Now
\(\Large \lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{dx}{1+cosx}\)
\(\Large =\lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{dx}{2cos^{2}\frac{x}{2}}=\Large \lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{1}{2}sec^{2}\frac{x}{2}dx\)
\(\Large =\lim\limits_{t\rightarrow 0}\left[tan\frac{x}{2} \right]_{0}^{\pi-t}\)
\(\Large =\lim\limits_{t\rightarrow 0}tan \left(\frac{ \pi }{2}-\frac{t}{2}\right) \)
\(\Large =\lim\limits_{t\rightarrow 0}cot\frac{t}{2}=\infty\)
Therefore the given integral is divergent.
\(\Large \int\limits_{0}^{x}\frac{dx}{1+cosx}\)
when \(x= \pi \) then \(\Large \frac{1}{1+cos \pi }=\infty\)
Now
\(\Large \lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{dx}{1+cosx}\)
\(\Large =\lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{dx}{2cos^{2}\frac{x}{2}}=\Large \lim\limits_{t\rightarrow 0}\int\limits_{0}^{ \pi -t}\frac{1}{2}sec^{2}\frac{x}{2}dx\)
\(\Large =\lim\limits_{t\rightarrow 0}\left[tan\frac{x}{2} \right]_{0}^{\pi-t}\)
\(\Large =\lim\limits_{t\rightarrow 0}tan \left(\frac{ \pi }{2}-\frac{t}{2}\right) \)
\(\Large =\lim\limits_{t\rightarrow 0}cot\frac{t}{2}=\infty\)
Therefore the given integral is divergent.
Part of solved Real Analysis questions and answers : >> Elementary Mathematics >> Real Analysis
Comments
Similar Questions
