What is the integral of sec 2x * tan 2x dx?

[tex]\int(\sec2x\cdot\tan2x)dx=\int\left(\dfrac{1}{\cos2x}\cdot\dfrac{\sin2x}{\cos2x}\right)dx\\\\=\int\dfrac{\sin2x}{\cos^22x}dx\Rightarrow\left|\begin{array}{ccc}\cos2x=t\\-2\sin2x\ dx=dt\\\sin2x\ dx=-\frac{1}{2}dt\end{array}\right|\Rightarrow\int\dfrac{-\frac{1}{2}}{t^2}dt\\\\=-\dfrac{1}{2}\int\dfrac{1}{t^2}dt=-\dfrac{1}{2}\int t^{-2}dt=-\dfrac{1}{2}\left(-t^{-1}\right)+C\\\\=\dfrac{1}{2t}+C=\boxed{\frac{1}{2\cos2x}+C}[/tex]

meerkat18 meerkat18 · Answer 2 · 2016-07-20T03:28:19+02:00

meerkat18 meerkat18

20-07-2016

We are asked to determine the integral of sec 2x tan 2x dx
sec 2x is equivalent to 1/cos2x while tan 2x is equal to sin 2x/cos 2x
In this case, the expression becomes sin 2x/ cos^2 2x

We use cos 2x as u so the du is equal to -2 sin 2x dx
The equation then becomes
-1/2∫1/u^2 du
=0.5/cos 2x + c = 0.5 sec 2x + c, where c is a constant

Answer Link