# Find the integrals \int\sec t(\sec t+\tan t)dt

coexpennan 2021-10-11 Answered
Find the integrals
$\int \mathrm{sec}t\left(\mathrm{sec}t+\mathrm{tan}t\right)dt$
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## Expert Answer

Alannej
Answered 2021-10-12 Author has 104 answers

Given,
$\int \mathrm{sec}\left(t\right)\left(\mathrm{sec}\left(t\right)+\mathrm{tan}\left(t\right)\right)dt$
Now,
$\therefore \int \mathrm{sec}\left(t\right)+\mathrm{tan}\left(t\right)\right)dt$
Let,
$\therefore A=\int \mathrm{sec}\left(t\right)\mathrm{tan}\left(t\right)dt$
Applying substitutuion method $u=\mathrm{sec}\left(t\right)$
$dy=\mathrm{tan}\left(t\right)dt$
$⇒A=\int 1du$
$⇒A=u+C$
substituting back: $u=\mathrm{sec}\left(t\right)$
$⇒A=\mathrm{sec}\left(t\right)+C$
The integral becomes,
$=\int {\mathrm{sec}}^{2}\left(t\right)dt+A\mathrm{sec}\left(t\right)$
$=\mathrm{tan}\left(t\right)+\mathrm{sec}\left(t\right)+C$
$\therefore \mathrm{sec}\left(t\right)\left(\mathrm{sec}\left(t\right)+\mathrm{tan}\left(t\right)dt=\mathrm{tan}\left(t\right)+\mathrm{sec}\left(t\right)+C$

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