11Bhaiya, even though u have taken the sum wrongly, the final answer is correct! [11]
How? [5][5][5][5][5]
7 Answers
620∫πxtanx dx/(secx + tanx) dx
=
0∫π x sinx dx/(1 + sinx) dx
=0∫π (π-x) sin(π-x) dx/(1 + sin(π-x)) dx
=0∫π (π-x) sinx dx/(1 + sinx) dx
taking the sum,
2I=0∫π πdx
2I=Ï€2
I=Ï€2/2
Check if there are mistakes :)
11Bhaiya, i think there are mistakes :(
Not so important mistake:
There are two dx in each step :P
Very important mistake:
While adding, remaining term will not be π but π sinx/(1+ sinx) :( (PLs correct me if i'm wrong)
11
11Arey but how is the final answer correct? [5][5][5][5][5][5]
11Here is the actual method:
Steps are same as bhaiya has done till that addition part
2I = π0∫πsinx dx/(sinx+1)
Multiply and divide by 1-sinx
2I = π0∫π(sinx-sin2x) dx/cos2x
= π[ 0∫π(sinx dx/cos2x - 0∫πsec2x-1 dx ]
on evaluating, we get
I = π2/2
The answer which Nishant bhaiya got by doing the sum wrongly [5]