How do you integrate $sec 2 x tan 2 x d x$ $sec2x tan2x dx$ ?

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How do you integrate $sec 2 x tan 2 x d x$ $sec2x tan2x dx$ ?

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Answer 1 · 2015-02-21T23:07:24

I would write the integral as:

$\int sec (2 x) tan (2 x) d x = \int \frac{1}{cos (2 x)} \frac{sin (2 x)}{cos (2 x)} d x =$ $intsec(2x)tan(2x)dx=int1/cos(2x)sin(2x)/cos(2x)dx=$

we can see that:

$d [cos (2 x)] = - 2 sin (2 x) d x$ $d[cos(2x)]=-2sin(2x)dx$
and so:
$sin (2 x) d x = \frac{d [cos (2 x)]}{- 2}$ $sin(2x)dx=(d[cos(2x)])/-2$

with this in mind I write the integral as:

$\int \frac{1}{{cos}^{2} (2 x)} \cdot \frac{1}{- 2} d [cos (2 x)] =$ $int1/cos^2(2x)*1/-2d[cos(2x)]=$
$= \frac{1}{- 2} \int {cos}^{- 2} (2 x) d [cos (2 x)] =$ $=1/-2intcos^(-2)(2x)d[cos(2x)]=$ (treating $cos (2 x)$ $cos(2x)$ as if it were $x$ $x$ in a normal integration where you get $\frac{x^{n + 1}}{n + 1}$ $x^(n+1)/(n+1)$ )
$= \frac{1}{2} \cdot \frac{1}{cos (2 x)} + c$ $=1/2*1/cos(2x)+c$

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How do you integrate $sec 2 x tan 2 x d x$ $sec2x tan2x dx$ ?

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