Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

can any body explain full solution step wise of 0 to "/2 integration of log sin x dx

can any body explain full solution step wise of



0 to "/2 integration of log sin x dx



Grade:Upto college level

1 Answers

Askiitians Expert Mohit Singla
19 Points
10 years ago

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

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free