Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int sec2x\tan 2x\mathrm{d}x$

Short answer

The solution of the given integral is $\int \sec 2x\tan 2x\mathrm{d}x=\frac{1}{2}\sec 2x+C$ .

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int sec2x\tan 2x\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=2x \\ \frac{du}{dx}=2 \\ d\mathrm{u}=2d\mathrm{x} \\ \frac{1}{2}d\mathrm{u}=d\mathrm{x} \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \sec 2x\tan 2x\mathrm{d}x=\frac{1}{2}\int \sec u\tan u\mathrm{du} \\ \int \sec 2x\tan 2x\mathrm{d}x=\frac{1}{2}\sec u+C \\ \int \sec 2x\tan 2x\mathrm{d}x=\frac{1}{2}\sec 2x+C \end{gathered}$$