Grade 12Integral Calculus
How to integrate root (1+secx)^1/2. Please mention full answer if possible
1 Answer
Ritika Das
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)∫u1/2.du.
=1/(sec x.tan x).(2/3).u3/2 + 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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