What's the integral of ∫sec⁡xtan⁡xdx?

burkinaval1b

burkinaval1b

Answered

2021-12-17

What's the integral of secxtanxdx?

Answer & Explanation

Medicim6

Medicim6

Expert

2021-12-18Added 33 answers

sec(x)tan(x)dx=sec(x)+C
Here is the explanation:
sec(x)tan(x)dx=1cos(x)×sin(x)cos(x)dx=sin(x)(cos(x))2dx=1cos(x)+C=sec(x)+C
Rita Miller

Rita Miller

Expert

2021-12-19Added 28 answers

u=sec(x)
du=(1cos(x))=1cos2(x)×(sin(x)dx)=sin(x)cos2(x)dx=tan(x)sec(x)dx
Thus,
tan(x)sec(x)dx=du=u=sec(x)+C
nick1337

nick1337

Expert

2021-12-28Added 573 answers

sec(x)tan(x)=d(sec(x))=sec(x)+C

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