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 you please explain question number 18 to me This is urgent

Can you please explain question number 18 to me
This is urgent
 
 

Question Image
Grade:12

1 Answers

Arun
25763 Points
2 years ago
I = ∫ 1/ √[ sin³ x. sin (x+a) ] dx 

= ∫ 1 / { sin x. √[ sin x.( sin x. cos a + cos x. sin a )] } dx 

= ∫ csc x / √[ sin² x.(cos a + (cos x / sin x). sin a) ] dx 

= ∫ csc x / [ sin x.√( cos a + cot x. sin a ) ] dx 

= ∫ csc² x / √( cos a + cot x. sin a ) dx ..................... (1) 
_________________________________ 

Let : u = cos a + cot x. sin a 

Then : du/dx = - sin a. csc² x 

so that : csc² dx = - csc a du. 
__________________________________ 

From (1), then, 

I = ∫ ( 1 / √u ) ( - csc a ) du 

= - csc a. ( 2√u ) + C 

= - 2. csc a. √( cos a + cot x. sin a ) + C

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