What is the integral of $(secxtanx)(1+secx)dx$ ?

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Answer 1 · 2015-03-24T00:01:33

$1/2 (1+secx)^2+C$

$int (secxtanx)(1+secx)dx=int (1+secx)(secxtanx)dx$

Let $u=1+secx$ . Then $du=secxtanxdx$

$int udu=1/2u^2+C$ So,

$int (secxtanx)(1+secx)dx=1/2(1+secx)^2+C$

What is the integral of (secxtanx)(1+secx)dx(secxtanx)(1+secx)dx?