How do you find the indefinite integral of $\int (sec t + tan t)$ $int (sect+tant)$ ?

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Answer 1 · 2018-02-05T16:16:08

The answer is $= ln (| tan t + sec t |) - ln (| cos t |) + C$ $=ln(|tant+sect|)-ln(|cost|)+C$

We calculate separately $\int sec t d t$ $intsectdt$ and $\int tan t d t$ $inttantdt$

$\int tan t d t = \int \frac{sin t d t}{cos t}$ $inttantdt=int(sintdt)/cost$

Perform the substitution

$u = cos t$ $u=cost$ , $\Rightarrow$ $=>$ , $d u = - sin t d t$ $du=-sintdt$

$\int tan t d t = \int \frac{- d u}{u} = - ln (u) = - ln (| cos t |)$ $inttantdt=int(-du)/u=-ln(u)=-ln(|cost|)$

$\int sec t = \int \frac{sec t (tan t + sec t) d t}{tan t + sec t}$ $intsect=int(sect(tant+sect)dt)/(tant+sect)$

Perform the substitution

$u = tan t + sec t$ $u=tant+sect$ , $\Rightarrow$ $=>$ , $d u = ({sec}^{2} t + sec t tan t) d t$ $du=(sec^2t+sect tant)dt$

$\int sec t = \int \frac{d u}{u} = ln u = ln (| tan t + sec t |)$ $intsect=int(du)/(u)=lnu=ln(|tant+sect|)$

Finally,

How do you find the indefinite integral of ∫(sect+tant)int (sect+tant)?