integration of 1/1+e ki power sin x??????

Asked by shashank tripathi | 20th Jun, 2013, 04:32: AM

Expert Answer:

This question can'e be of indefinite integrals. It should be a question of definite integrals with limits defined from 0 to 2pi. 
 
Then, the solution would be 
 

Let I =limit 0 to 2pi integral 1dx / 1+esinx _______(1)

Now , as limit 0 to a integral f(x)dx = limit 0 to a integral f(a-x)dx

So , I = integral dx /1+e-sinx

I = integral esinx dx / 1+esinx _______(2)

Adding (1) & (2)

2I = limit 0 to 2pi integral (1+esinx)dx /1+esinx

2I = limit 0 to 2pi integral 1dx

2I = limit 0 to 2pi [x]

2I = 2pi – 0

I = pi

Answered by  | 20th Jun, 2013, 05:30: AM

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