Answered question

2022-05-12

Answer & Explanation

Nick Camelot

Nick Camelot

Skilled2023-05-07Added 164 answers

To solve secudu, we use the substitution v=tanu+secu. Then, dv/dx=sec2u+tanusecu=secu(tanu+secu)=secu(v). Thus, du=dv/secu(v), and
secudu=secusecu(tanu+secu)dv
=1vdv
=ln|v|+c
=ln|tanu+secu|+c.
Therefore,
secudu=ln|tanu+secu|+c.
We can check this result by differentiating both sides with respect to u and using the identity ddxln|x|=1x:
ddu(ln|tanu+secu|+c)&=1tanu+secu·(secutanu+sec2u)
=1tanu+secu·secu(tanu+secu)
=secu,
which is the integrand we started with. Thus, our result is correct.

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