If ∫(1+√tanx)/sin2xdx=Alogtanx+B√tanx+C then 4A-B=

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MARKS Content Team · Accepted Answer

Correct answer: (C) 1. I=∫(1+√tanx)/sin2xdx=∫1/sin2x+√tanxsin2xdx =∫1/2sinxcosxdx+∫√sinx2sinxcosx√cosxdx =∫1/2sinxcosxcos^2xdx+∫1/(2√sinx√cosxcos^2x)dx =∫(sec^2x)/2tanxdx+∫(sec^2x)/(2√tanx)dx =1/2lntanx+√tanx+C So, 4A-B=4×1/2-1=1

If $\int \frac{1 + \sqrt{\tan x}}{\sin 2 x} d x = A logtan x + B \sqrt{\tan x} + C$ then $4 A - B =$

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Relevant Questions

If ∫1+tanxsin2xdx=Alogtanx+Btanx+C then 4A-B=

Solution

Relevant Questions

If $\int \frac{1 + \sqrt{\tan x}}{\sin 2 x} d x = A logtan x + B \sqrt{\tan x} + C$ then $4 A - B =$