Definite Integral

∫pi011+tanxdx

12 Answers

1
venkateshan ·

pi/2

1
sidharthkumar ·

zero

1
Vinay Arya ·

Please Please Please Please.Show me the working.

39
Dr.House ·

the indefinite integral is

(1/2) (x + ln| sin x + cos x |)

evaluating it between 0 and pi u get pi/2 as the answer

1
Vinay Arya ·

I asked for working not the direct answer.

1
Vinay Arya ·

And yes ,the answer is correct.It is indeed pi/2.

39
Dr.House ·

http://in.answers.yahoo.com/question/index?qid=20100421045250AAFKPYc

1
Vinay Arya ·

Thank you very much.The first method is quite great.

341
Hari Shankar ·

\int_0^{\pi} \frac{dx}{1+\tan x} = \frac{1}{2} \int_0^{\pi} 1 + \frac{1-\tan x}{1+ \tan x} \ dx= \frac{\pi}{2} + \frac{1}{2} \int_0^{\pi} \tan \left(\frac{\pi}{4}-x \right) \ dx

\int_0^{\pi} \tan \left(\frac{\pi}{4}-x \right) \ dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \tan y \ dy = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot y \ dy = 0

Hence the integral evaluates to

\frac{\pi}{2}

1
redox ·

but tan x is discontinous at π/2,we have to break the limits in the last step

39
Pritish Chakraborty ·

lol! Would you believe it....the Yahoo Answers page !^--_--^! posted...the second answer on it is by me xD

341
Hari Shankar ·

@redox - not if you are invoking a result about integral of an odd function about an interval symmetric about x=0.

Your Answer

Close [X]