Secx+[(tanx)^2/(1+secx)]=?

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Satyanarayana Rao M

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Sec x + = Sec x + =Sec x + = Sec x + Sec x - 1 = 2 Sec x -1
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sec x + =sec x + since 1+ (tan x)^2 =(sec x)^2 sec x +=sec x +(sec x - 1)= 2 sec x - 1
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Note:-{2secx-1} and there is a restriction on 'x'. x not equal to(2n+1)pi.
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2(Secx) -1
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Secx+ = Secx+ = Secx+ = Secx+ = 1
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Strategy: The strategy is not obvious. Multiply and divide by (sec x + tan x); use Substitution. (integral) sec x dx = (integral) sec x sec x + tan x sec x + tan x dx set u = sec x + tan x then we find du = (sec x tan x + sec2 x) dx substitute du = (sec x tan x + sec2 x) dx, u =...
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Strategy: The strategy is not obvious. Multiply and divide by (sec x + tan x); use Substitution. (integral) sec x dx = (integral) sec x sec x + tan x sec x + tan x dx set u = sec x + tan x then we find du = (sec x tan x + sec2 x) dx substitute du = (sec x tan x + sec2 x) dx, u = sec x + tan x (integral) sec x sec x + tan x sec x + tan x dx = (integral) (sec2 x + sec x tan x) dx sec x + tan x = (integral) du u solve integral = ln |u| + C substitute back u=sec x + tan x = ln |sec x + tan x| + C read less
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2secx-1
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Sec2x - Secx
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2Secx-1
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