MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
bharat bhushan sharma Grade: Upto college level
        

can any body explain full solution step wise of



0 to "/2 integration of log sin x dx



7 years ago

Answers : (1)

Askiitians Expert Mohit Singla
19 Points
										

Dear Bharat,


I=∫ log(sin x).dx         .............eq 1
I=∫logsin(π/2-x)dx
I=∫logcosx dx            ..................eq2
adding equation 1&2
2I=∫logsinx+logcosx dx
=∫logsinx*cosx dx
=∫log(sin2x)/2 dx
=∫[logsin2x-log2] dx
=∫logsin2xdx - ∫log2dx .................eq 3


therefore 2I=A-B


where,A=∫logsin2xdx
and B=∫log2dx
with limits from 0 to π/2
Solving B,we get B=log2[x],with limits from 0 to π/2
or B=π/2log2


in the first integral A put 2x=t
d.w.t.x
dx=dt/2
where x=0,t=0 where x=π/2 ,t=π
∫logsint dt with limit t= o & t=π
now
2I=1/2∫logsint dt-B
Now applying property



Here F(2a-x)=F(x)


Therefore 2I=1/2.2∫logsint dt-B with limits from 0 to π/2
2I=I-B


I=-B=-π/2log2


or I=π/2log(1/2).



Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.



All the best Bharat !!!


 



Regards,


Askiitians Experts


MOHIT


 

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete AIPMT/AIIMS Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 3,180 off
USE CODE: CART20
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 3,180 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details