What is $\int sec (2 x) tan (2 x) d x$ $intsec(2x)tan(2x) dx$ ?

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Answer 1 · 2015-12-31T06:19:28

$\frac{1}{2} sec (2 x) + C$ $1/2 sec(2x) + C$

$\int sec (2 x) tan (2 x) d x$ $int sec(2x)tan(2x) dx$
Let $u = 2 x$ $u = 2x$
$d u = 2 d x$ $du = 2dx$
$(d \frac{u}{2}) = d x$ $(du/2)=dx$

The integral is rewritten as

$\int sec (u) tan (u) \frac{d u}{2}$ $int sec(u)tan(u) (du)/2$
$= \frac{1}{2} \int sec (u) tan (u) d u$ $=1/2 int sec(u)tan(u) du$

$= \frac{1}{2} sec (u) + C$ $=1/2 sec(u) + C$

Substituting back for $u$ $u$ we get $\frac{1}{2} sec (2 x) + C$ $1/2 sec(2x) + C$

What is intsec(2x)tan(2x) dx∫sec(2x)tan(2x)dxintsec(2x)tan(2x) dx?