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Install NowAdarsh Udayan
Feb 11, 2015
Adarsh Udayan
Jan 20, 2015
Let I =
Shreedhar Chaudhury
Aug 2, 2014
(1 / Sin (b -a) ∫ Sin [ (x - a) - (x - b) ] dx / Sin (x - a) Sin (x - b)
(1 / Sin (b -a) { ∫ [Cos (x - a) Sin(x - b) - Sin(x - a) Cos (x - b) ] dx / Sin (x - a) Sin (x - b)} .<...
Shreedhar Chaudhury
Aug 4, 2014
Given ∫ dx / Cos (x-a) Cos (x-b)
⇒ (1 / Sin (b -a) ∫ Sin [ (x - a) - (x - b) ] dx / Cos (x-a) Cos (x-b)
⇒ (1 / Sin (b -a) { ∫ [Cos (x...
Shreedhar Chaudhury
Aug 5, 2014
Given ∫ dx / Sin (x-a) Cos (x-b)
⇒ (1 / Cos (b -a) ∫ Cos [ (x - a) - (x - b) ] dx / Sin (x - a) Cos (x - b)
⇒ (1 / Cos (b -a) { ∫ [Cos...
Divya
Jul 13, 2014
Siddharth
Oct 26, 2013
Anubhav Singh
Feb 10, 2015
Raghu Mishra
May 23, 2015
Priya
Oct 5, 2013
Adarsh Udayan
Feb 11, 2015
Adarsh Udayan
Jan 20, 2015
Let I =
Shreedhar Chaudhury
Aug 2, 2014
(1 / Sin (b -a) ∫ Sin [ (x - a) - (x - b) ] dx / Sin (x - a) Sin (x - b)
(1 / Sin (b -a) { ∫ [Cos (x - a) Sin(x - b) - Sin(x - a) Cos (x - b) ] dx / Sin (x - a) Sin (x - b)} .<...
Shreedhar Chaudhury
Aug 4, 2014
Given ∫ dx / Cos (x-a) Cos (x-b)
⇒ (1 / Sin (b -a) ∫ Sin [ (x - a) - (x - b) ] dx / Cos (x-a) Cos (x-b)
⇒ (1 / Sin (b -a) { ∫ [Cos (x...
Shreedhar Chaudhury
Aug 5, 2014
Given ∫ dx / Sin (x-a) Cos (x-b)
⇒ (1 / Cos (b -a) ∫ Cos [ (x - a) - (x - b) ] dx / Sin (x - a) Cos (x - b)
⇒ (1 / Cos (b -a) { ∫ [Cos...
Divya
Jul 13, 2014
Siddharth
Oct 26, 2013
Anubhav Singh
Feb 10, 2015
Raghu Mishra
May 23, 2015
Priya
Oct 5, 2013