Question

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

Answer 1

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

Answer 2

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