How to integrate root (1+secx)^1/2. Please mention full answer if possible

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Ritika Das · Answer

Let u = 1+sec x. du/dx=sec x.tan x. dx=(1/sec x.tan x).du. Now, I=∫sqrt.(1+sec x).dx. Putting values from above we get: – =∫sqrt.(u).du.(1/(sec x.tan x)). =1/(sec x.tan x)∫u 1/2 .du. =1/(sec x.tan x).(2/3).u 3/2 + c. Putting back value of u we get the final answer: – I = 2/(3sec x.tan x).(1+sec x) 3/2 + c. Please approve my answer if you like the explanation!

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How to integrate root (1+secx)^1/2. Please mention full answer if possible

How to integrate root (1+secx)^1/2. Please mention full answer if possible

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