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Grade: 12
        
Evaluate this question integral of {cotx/log(sinx)dx}?
8 months ago

Answers : (2)

Arun
13827 Points
							
Dear student
 

 I = ⌡ {cotx / log(sinx) } dx

Let u=log(sinx)

du/dx = (1/sinx) * d(sinx)/dx

du/dx = cosx/sinx

du = cotx dx

Therefore : I = ⌡ {1 / u } du

I = log(u) + C

I = log(logIsinxI)+C

 

Regards

Arun (askIITians forum expert)

8 months ago
Shubham Kumar
20 Points
							Whenever this type of integration comes it has following unique generalised resultIntegral of f`(x)/f(x)=log(modulus of f(x))Where f`(x)= differentiation of f(x)
						
8 months ago
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